Answer:
Decreasing the cross-sectional area of the wire will increase the resistance of the wire.
Increasing the resistivity of the material the wire is composed of will increase the resistance of the wire.
Increasing the length of the wire will increase the resistance of the wire.
Explanation:
The resistance of a piece of wire is given by:
[tex]R=\rho \frac{L}{A}[/tex]
where
[tex]\rho[/tex] is the resistivity of the material
L is the length of the wire
A is the cross-sectional area of the wire
We notice that:
- The resistance of the wire is directly proportional to the resistivity of the material and the to the length of the wire
- The resistance of the wire is inversely proportional to the cross-sectional area of the wire
This means the following are correct:
Decreasing the cross-sectional area of the wire will increase the resistance of the wire.
Increasing the resistivity of the material the wire is composed of will increase the resistance of the wire.
Increasing the length of the wire will increase the resistance of the wire.
Light has been argued as being a wave and then a particle, and back again. Which of the following is right? 1. Light’s true nature is as a wave. 2. Light’s true nature lies somewhere in between wave and particle models. 3. Light’s true nature is as a particle. 4. Light’s true nature is wave-particle duality.
Both light and matter can possess particle properties and wave properties at the same time.
This quantum behavior was proposed by the French physicist Louis De Broglie in 1924 in his doctoral thesis when proposing that all matter has an associated wave, and light did not escape from that behavior. Then this was empirically tested in various experiments where light behaved as a photon (particle) and in others as particles.
Einstein also demonstrated this fact, however, in spite of being observed this behavior several times, it could not have been done simultaneously, until in 2015 a group of scientists from the Federal Polytechnic School of Lausanne (Switzerland), succeeded to perform an experiment and capture for the first time the image of the wave-particle duality of light.
You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.8 × 107 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits 4.6 × 1011 m from its star with a period of 402 earth days. Once on the surface you find that the acceleration due to gravity is 22.4 m/s2. What are the mass of (a) the planet and (b) the star? (G = 6.67 × 10-11 N · m2/kg2) You are the science officer on a visit to a distant solar system. Prior to landing on a planet you measure its diameter to be 1.8 × 107 m and its rotation period to be 22.3 hours. You have previously determined that the planet orbits 4.6 × 1011 m from its star with a period of 402 earth days. Once on the surface you find that the acceleration due to gravity is 22.4 m/s2. What are the mass of (a) the planet and (b) the star? (G = 6.67 × 10-11 N · m2/kg2) (a) 2.7 × 1025 kg, (b) 2.8 × 1031 kg (a) 5.0 × 1025 kg, (b) 2.8 × 1031 kg (a) 5.0 × 1025 kg, (b) 4.8 × 1031 kg (a) 2.7 × 1025 kg, (b) 4.8 × 1031 kg
(a) [tex]2.7\cdot 10^{25} kg[/tex]
The acceleration due to gravity on the surface of the planet is given by
[tex]g=\frac{GM}{R^2}[/tex] (1)
where
G is the gravitational constant
M is the mass of the planet
R is the radius of the planet
Here we know:
[tex]g=22.4 m/s^2[/tex]
[tex]d=1.8\cdot 10^7 m[/tex] is the diameter, so the radius is
[tex]R=\frac{d}{2}=\frac{1.8\cdot 10^7 m}{2}=9\cdot 10^6 m[/tex]
So we can re-arrange eq.(1) to find M, the mass of the planet:
[tex]M=\frac{gR^2}{G}=\frac{(22.4 m/s^2)(9\cdot 10^6 m)^2}{6.67\cdot 10^{-11}}=2.7\cdot 10^{25} kg[/tex]
(b) [tex]4.8\cdot 10^{31}kg[/tex]
The planet is orbiting the star, so the centripetal force is equal to the gravitational attraction between the planet and the star:
[tex]m\frac{v^2}{r}=\frac{GMm}{r^2}[/tex] (1)
where
m is the mass of the planet
M is the mass of the star
v is the orbital speed of the planet
r is the radius of the orbit
The orbital speed is equal to the ratio between the circumference of the orbit and the period, T:
[tex]v=\frac{2\pi r}{T}[/tex]
where
[tex]T=402 days = 3.47\cdot 10^7 s[/tex]
Substituting into (1) and re-arranging the equation
[tex]m\frac{4\pi r^2}{rT^2}=\frac{GMm}{r^2}\\\frac{4\pi r}{T^2}=\frac{GM}{r^2}\\M=\frac{4\pi r^3}{GMT^2}[/tex]
And substituting the numbers, we find the mass of the star:
[tex]M=\frac{4\pi^2 (4.6\cdot 10^{11} m)^3}{(6.67\cdot 10^{-11})(3.47\cdot 10^7 s)^2}=4.8\cdot 10^{31}kg[/tex]
A major motor company displays a die-cast model of its first automobile, made from 9.00kg of iron. To celebrate its hundredth year in business, a worker will recast the model in gold from the original dies. What mass of gold is needed to make the new model? The density of iron is 7.86 ✕ 10^3 kg/m3, and that of gold is 19.30 ✕ 10^3 kg/m3
Answer:
22.10 kg
Explanation:
The worker wants to recast the model from iron to gold, so the volume of the model made of iron and made of gold will be the same:
[tex]V_i = V_g[/tex]
the volume of each model can be rewritten as the ratio between the mass of the model, m, and the density of the material, d:
[tex]V=\frac{m}{d}[/tex]
so we can rewrite the first equation as
[tex]\frac{m_i}{d_i}=\frac{m_g}{d_g}[/tex]
where we have
[tex]m_i = 9.00 kg[/tex] is the mass of the model in iron
[tex]d_i = 7.86\cdot 10^3 kg/m^3[/tex] is the density of iron
[tex]m_g[/tex] is the mass of gold needed
[tex]d_g = 19.30\cdot 10^3 kg/m^3[/tex] is the density of gold
Solving for [tex]m_g[/tex], we find
[tex]m_g = d_g \frac{m_i}{d_i}=(19.30\cdot 10^3 kg/m^3) \frac{9.00 kg}{7.86\cdot 10^3 kg/m^3}=22.10 kg[/tex]
Can you cool a kitchen by leaving the refrigerator door open while the refrigerator is operating? a. No. Since the refrigerator is operating, the “heat pump” transfers thermal energy from the inside of the refrigerator to the outside of the refrigerator which is presumably in the same room. Since this process is not 100% efficient due to the second law, the average temperature of the room will increase over time. b. No. Since the refrigerator is operating, the “heat pump” transfers thermal energy from the inside of the refrigerator to the outside of the refrigerator which is presumably in the same room. Due to the second law, there will be no change in temperature due to the perfect efficiency of the heat pump.
Answer:
no
Explanation:
because will the cool air will be leaving the fridge the motor in the bottom will be running and eventually the mixture of the cool and hot air will balance out
'a' is the correct statement.
If a negatively-charged rod is brought close to an electroscope that is negatively charged, what will the two leaves of the electroscope do?
move closer together
move farther apart
not move at all
become positively charged
Answer:
Move further apart
Explanation:
The negative charge on the rod repels the electrons on the cap thus they concentrate at the leaves causing an increase in the amount of charge at the leaves. Since like charges repel, the two leaves of the electroscope move further apart.
Final answer:
When a negatively-charged rod is brought near a negatively-charged electroscope, the leaves of the electroscope will move farther apart due to increased electrostatic repulsion.
Explanation:
When a negatively-charged rod is brought close to an already negatively-charged electroscope, the electrostatic repulsion between the like charges will cause the two leaves of the electroscope to move farther apart. This is due to an increase in the density of negative charges within the leaves of the electroscope, which causes the leaves to repel each other even more.
Charging by induction occurs when a charged object is brought near, but not in contact with, a conductor. However, since the electroscope in this case is already negatively charged, bringing another negatively charged object close will simply enhance the repulsion between charges.
can someone help me?!!!!!
Answer:
magnitude: 21.6; direction: 33.7 degrees
Explanation:
When we multiply a vector by a scalar, we have to multiply each component of the vector by the scalar number. In this case, we have
vector: (-3,-2)
Scalar: -6
so the vector multiplied by the scalar will have components
[tex](-3\cdot (-6), -2 \cdot (-6))=(18,12)[/tex]
The magnitude is given by Pythagorean's theorem:
[tex]m=\sqrt{18^2+12^2}=21.6[/tex]
and the direction is given by the arctan of the ratio between the y-component and the x-component:
[tex]\theta = tan^{-1} (\frac{12}{18})=33.7^{\circ}[/tex]
can someone help me?!!!!!
Answer:magnitude -5; angle 160°
Explanation:
Vector A is described as having magnitude 5 and angle -20°.
To get an equivalent vector, we either leave the magnitude at 5 and add 360° to the angle, or we reverse the magnitude to -5 and add 180° to the angle.
5 @ -20° = 5 @ 340°
5 @ -20° = -5 @ 160°
The third one is the answer.
On coordinate axes, point A = (1.50, 2.00), measured in meters. Let the voltage value very far away (“infinity”) be defined as zero (i.e. V∞ = 0). To make things simpler, just express your answers in terms of e and k (in other words, don’t plug in numerical values for e and k). A lone proton is placed at the origin: a. Calculate the electric field (magnitude and direction) it will cause at point A. b. Calculate the magnitude and direction of the electric force it will exert on a proton located at point A. c. Calculate the magnitude and direction of the electric force it will exert on an electron located at point A. d. Calculate the uniform electric field (magnitude and direction) that would need to be present (in addition to the proton), so that an electron located at point A would have no acceleration. e. Calculate the electric potential (i.e. the voltage) that the lone proton will cause at point A. f. Calculate the electric potential energy (UE) that a second proton would have if located at point A. g. Calculate the electric potential energy (UE) that an electron would have if located at point A.
a. [tex]E=k\frac{e}{6.25}[/tex], [tex]53.1^{\circ}[/tex] above x-axis
The electric field produced by a single point charge is
[tex]E=k\frac{q}{r^2}[/tex]
where
q is the charge
r is the distance from the charge
In this case,
q = e
the charge is at the origin, so the distance of point A (1.50, 2.00) from the charge is
[tex]r=\sqrt{(1.50-0)^2+(2.00-0)^2}=2.5[/tex]
So the magnitude of the electric field at point A is
[tex]E=k\frac{e}{(2.5)^2}=k\frac{e}{6.25}[/tex]
The electric field produced by a single-point positive charge is radial, pointing away from the charge. So the direction of the field in this case is given by
[tex]\theta = tan^{-1} (\frac{A_y}{A_x})=tan^{-1} (\frac{2.00}{1.50})=53.1^{\circ}[/tex]
above the x-axis.
b. [tex]F=k\frac{e^2}{6.25}[/tex], [tex]53.1^{\circ}[/tex] above x-axis
The magnitude of the electric force exerted on a charge is given by
[tex]F=qE[/tex]
where
q is the charge magnitude
E is the electric field strength
In this case, we have a proton (charge e), and the electric field strength at point A is
[tex]E=k\frac{e}{6.25}[/tex]
So the force exerted on the proton is
[tex]F=(e)(k\frac{e}{6.25})=k\frac{e^2}{6.25}[/tex]
Moreover, the direction of the electric force exerted on a positive charge is equal to the direction of the electric field at that point, so the direction of the electric force is also [tex]53.1^{\circ}[/tex] above x-axis.
c. [tex]F=k\frac{e^2}{6.25}[/tex], [tex]233.1^{\circ}[/tex]
The electric force exerted on a charge is given by
[tex]F=qE[/tex]
In this case, we have an electron (charge e), and the electric field strength at point A is
[tex]E=k\frac{e}{6.25}[/tex]
So the force exerted on the electron is equal as the one for the proton:
[tex]F=(e)(k\frac{e}{6.25})=k\frac{e^2}{6.25}[/tex]
However, this time the direction is different: in fact, for a negative charge the electric force points in a direction opposite to the electric field. So, the direction of the electric force in this case is
[tex]\theta = 53.1^{\circ}+180^{\circ}=233.1^{\circ}[/tex]
d. [tex]E'=k\frac{e}{6.25}[/tex], [tex]233.1^{\circ}[/tex]
In order for the electron at point A not to have acceleration, the net force on it should be zero:
[tex]F_{net} =0[/tex]
The net force is the algebraic sum of the force calculated at point c), due to the electric field produced by the proton, and the additional force F' due to the uniform electric field that we need to add:
[tex]F_{net}=F+F' = 0\\F' = -F[/tex]
Rewriting [tex]F'=eE'[/tex]
where E' is the uniform electric field that we add, we find
[tex]E' = -\frac{F}{e}=-\frac{1}{e}k\frac{e^2}{6.25}=-k\frac{e}{6.25}\\[/tex]
So the magnitude of the uniform electric field must be
[tex]E'=k\frac{e}{6.25}[/tex]
and the direction must be opposite to the electric field produced by the proton, so at an angle of [tex]233.1^{\circ}[/tex].
e. [tex]V=k \frac{e}{2.5}[/tex]
The electric potential generated by a single-point positive charge is
[tex]V=k \frac{q}{r}[/tex]
where here we have
q = e is the charge of the proton
r = 2.5 is the distance of point A from the proton
Substituting into the formula, we find
[tex]V=k \frac{e}{2.5}[/tex]
f. [tex]k\frac{e^2}{2.5}[/tex]
The electric potential energy of a charge in an electric field is
[tex]U=qV[/tex]
where q is the charge and V is the electric potential
Here we have a proton (charge q=e) located at point A, where the potential is
[tex]V=k \frac{e}{2.5}[/tex]
So its electric potential energy is
[tex]U=(e)(k\frac{e}{2.5})=k\frac{e^2}{2.5}[/tex]
g. [tex]-k\frac{e^2}{2.5}[/tex]
As before, electric potential energy of a charge in an electric field is
[tex]U=qV[/tex]
Here we have an electron (charge q=-e) located at point A, where the potential is
[tex]V=k \frac{e}{2.5}[/tex]
So its electric potential energy this time is
[tex]U=(-e)(k\frac{e}{2.5})=-k\frac{e^2}{2.5}[/tex]
So it's the same as the one calculated at part f), but with a negative sign, since the charge is negative.
The initial temperature of 150 g of ice is ????20°C. The spe- cific heat capacity of ice is 0.5 cal/g·C° and water’s is 1 cal/g·C°. The latent heat of fusion of water is 80 cal/g. a. How much heat is required to raise the ice to 0°C and completely melt the ice? b. How much additional heat is required to heat the water (obtained by melting the ice) to 25°C? c. What is the total heat that must be added to convert the 80 g of ice at ????20°C to water at ????25°C? d. Can we find this total heat simply by computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 80 g of water by 45°C? Explain.
1. 13,500 cal
First of all, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. This is given by
[tex]Q_1 = m C_i \Delta T[/tex]
where
m = 150 g is the mass of the ice
C_i = 0.5 cal/g·C° is the specific heat capacity of the ice
[tex]\Delta T=0 C-(-20 C)=20^{\circ}C[/tex] is the change in temperature of the ice
Substituting,
[tex]Q_1 = (150 g)(0.5 cal/gC)(20 C)=1500 cal[/tex]
Now we have to find the amount of heat needed to melt the ice, which is
[tex]Q_2 = m \lambda_f[/tex]
where
m = 150 g is the mass of the ice
[tex]\lambda_f = 80 cal/g[/tex] is the latent heat of fusion
Substituting,
[tex]Q_2 = (150 g)(80 cal/g)=12,000 cal[/tex]
So the total heat required is
[tex]Q_3 = 1500 cal + 12,000 cal = 13,500 cal[/tex]
2. 3750 cal
The additional amount of heat required to heat the water to 25°C is
[tex]Q_4 = m C_w \Delta T[/tex]
where
m = 150 g is the mass of water
C_w = 1 cal/g·C is the speficic heat capacity of water
[tex]\Delta T=25 C-0 C=25^{\circ}C[/tex] is the change in temperature
Substituting,
[tex]Q_4 = (150 g)(1 cal/gC)(25 C)=3,750 cal[/tex]
3. 9200 cal
First of all, we need to find the amount of heat needed to raise the temperature of the ice from -20°C to 0°C. As at point 1., this is given by
[tex]Q_1 = m C_i \Delta T[/tex]
where
m = 80 g is the mass of the ice
C_i = 0.5 cal/g·C° is the specific heat capacity of the ice
[tex]\Delta T=0 C-(-20 C)=20^{\circ}C[/tex] is the change in temperature of the ice
Substituting,
[tex]Q_1 = (80 g)(0.5 cal/gC)(20 C)=800 cal[/tex]
Now we have to find the amount of heat needed to melt the ice:
[tex]Q_2 = m \lambda_f[/tex]
where
m = 80 g is the mass of the ice
[tex]\lambda_f = 80 cal/g[/tex] is the latent heat of fusion
Substituting,
[tex]Q_2 = (80 g)(80 cal/g)=6,400 cal[/tex]
Finally, the amount of heat required to heat the water to 25°C is
[tex]Q_3 = m C_w \Delta T[/tex]
where
m = 80 g is the mass of water
C_w = 1 cal/g·C is the speficic heat capacity of water
[tex]\Delta T=25 C-0 C=25^{\circ}C[/tex] is the change in temperature
Substituting,
[tex]Q_3 = (80 g)(1 cal/gC)(25 C)=2,000 cal[/tex]
So the total heat required is
[tex]Q=Q_1+Q_2+Q_3=800 cal+6,400 cal+2,000 cal=9,200 cal[/tex]
4. No
Explanation:
The total heat required for this process consists of 3 different amounts of heat:
1- The heat required to bring the ice at melting temperature
2- The heat required to melt the ice, while its temperature stays constant
3- The heat required to raise the temperature of the water
However, computing how much heat is required to melt the ice and adding the amount of heat required to raise the temperature of 80 g of water by 45°C is not equivalent: in fact, the calculation of point 1) requires to use the specific heat capacity of ice, not that of water, therefore the two are not equivalent.
A mass spectrograph is operated with deuterons, which have a charge of +e and a mass of 3.34 x 10-27 kg. Deuterons emerge from the source, which is grounded with negligible velocity. The velocity of the deuterons as they pass through the accelerator grid is 8.0 x 105 m/s. A uniform magnetic field of magnitude B = 0.20 T, directed out of the plane, is present at the right of the grid. In the figure, the deuterons are in circular orbit in the magnetic field. The radius of the orbit and the initial sense of deflection are closest to:
Answer:
0.0835 m
Explanation:
The magnetic force exerted on the deuteron is equal to the centripetal force that keeps it in circular motion:
[tex]qvB = m\frac{v^2}{r}[/tex]
where
[tex]q=+e = 1.6\cdot 10^{-19} C[/tex] is the charge
[tex]v=8.0\cdot 10^5 m/s[/tex] is the speed of the deuteron
[tex]B=0.20 T[/tex] is the magnetic field strength
[tex]m=3.34\cdot 10^{-27} kg[/tex] is the mass of the deuteron
r is the radius of the orbit
re-arranging the equation, we find:
[tex]r=\frac{mv}{qB}=\frac{(3.34\cdot 10^{-27}kg)(8.0\cdot 10^5 m/s)}{(1.6\cdot 10^{-19} C)(0.20 T)}=0.0835 m[/tex]
An object of mass 10 kg is released at point A, slides to the bottom of the 30° incline, then collides with a horizontal massless spring, compressing it a maximum distance of 0.75 m. The spring constant is 500 M/m, the height of the incline is 2.0 m, and the horizontal surface is frictionless. (a) What is the speed of the object at the bottom of the incline? (b) What is the work of friction on the object while it is on the incline? (c) The spring recoils and sends the object back toward the incline. What is the speed of the object when it reaches the base of the incline?
Answer:
Explanation:
The energy stored in the spring = the kinetic energy at the bottom of the incline
1/2 kx² = 1/2 mv²
kx² = mv²
(500 N/m) (0.75 m)² = (10 kg) v²
v ≈ 5.3 m/s
The energy stored in the spring = the initial potential energy - work done by friction
1/2 kx² = mgh - W
1/2 (500 N/m) (0.75 m)² = (10 kg) (9.8 m/s²) (2.0 m) - W
W ≈ 55 J
Since the horizontal surface is frictionless, the object will have the same speed at the bottom of the incline as before.
v ≈ 5.3 m/s
Answer:
In this problem we need to use the conservation of energy theorem, which is about to constant exchange between kinetic and potential energy. In this case, we have two important points to analyse. The first one is at the initial point, where the mass is just released at 2 m from the ground. The final point would be when the mass hits the massless spring.
So, in the initial point the total energy of the mass would be only potential energy due to its position and its lack of speed. Then, the mass will be released gaining speed, until reaches the spring, where all the kinetic energy is transformed in potential energy but due to elastic forces (spring). So, in the final point, both energy are the same, the kinetic and the elastic potential energy, which we express like this:
[tex]K_{f}=U_{f}[/tex]
But, the definition of each one is:
[tex]K=\frac{1}{2}mv^{2} \\U=\frac{1}{2}kx^{2}[/tex]
Where k is the constant of the spring. So, we replace this in the initial equation and solve for v:
[tex]\frac{1}{2}mv_{f} ^{2}=\frac{1}{2}kx^{2}\\v_{f}=\sqrt{\frac{kx^{2}}{m}}\\v_{f}=\sqrt{\frac{500(0.75)^{2} }{10}}=5.30m/s[/tex]
Therefore, the speed in the spring point is 5.30 m/s.
Now, to calculate the work due to friction, we have to analyse first, because the problem doesn't offer a friction coefficient, but, we can analyse according to energies again, because work and energy are magnitudes close related, and they have the same dimension.
So, if we thought about it, we'll find that the work due to friction is what slows down the mass, subtracting its potential energy from the initial point, giving as a result the final potential energy:
[tex]U_{f}=U_{i}-W_{k}[/tex]
[tex]\frac{1}{2}kx^{2} =mgy-W_{k} \\W_{k}=10(9.8)(2)-\frac{1}{2}(500)(0.75)^{2}\\W_{k}=196-140.625=55.38J[/tex]
Therefore, the word due to friction is 55.38 J.
Lastly, the option c is asking about the speed of the object when it reaches the vase of the incline, which is the horizontal ground. The problem specify that this portion of the trajectory is frictionless, which mean that nothing slows down the mass.
Therefore, the speed in the horizontal part is the same 5.30 m/s.
The Schwarzschild radius is the distance from an object at which the escape velocity is equal to the speed of light. A black hole is an object that is smaller than its Schwarzschild radius, so not even light itself can escape a black hole. The Schwarzschild radius ???? depends on the mass ???? of the black hole according to the equation ????=2????????????2 where ???? is the gravitational constant and ???? is the speed of light. Consider a black hole with a mass of 5.7×107M⊙. Use the given equation to find the Schwarzschild radius for this black hole. Schwarzschild radius: m What is this radius in units of the solar radius
1. [tex]1.69\cdot 10^{11} m[/tex]
The Schwarzschild radius of an object of mass M is given by:
[tex]r_s = \frac{2GM}{c^2}[/tex] (1)
where
G is the gravitational constant
M is the mass of the object
c is the speed of light
The black hole in the problem has a mass of
[tex]M=5.7\cdot 10^7 M_s[/tex]
where
[tex]M_s = 2.0\cdot 10^{30} kg[/tex] is the solar mass. Substituting,
[tex]M=(5.7\cdot 10^7)(2\cdot 10^{30}kg)=1.14\cdot 10^{38} kg[/tex]
and substituting into eq.(1), we find the Schwarzschild radius of this black hole:
[tex]r_s = \frac{2(6.67\cdot 10^{-11})(1.14\cdot 10^{38} kg)}{(3\cdot 10^8 m/s)^2}=1.69\cdot 10^{11} m[/tex]
2) 242.8 solar radii
We are asked to find the radius of the black hole in units of the solar radius.
The solar radius is
[tex]r_S = 6.96\cdot 10^5 km = 6.96\cdot 10^8 m[/tex]
Therefore, the Schwarzschild radius of the black hole in solar radius units is
[tex]r=\frac{1.69\cdot 10^{11} m}{6.96\cdot 10^8 m}=242.8[/tex]
The Stanford Linear Accelerator Center (SLAC) accelerates electrons such that the 3 km proper length tube appears to be only 8 cm long due to Lorentz contraction. a. Find the values of gamma and the electrons’ speed. b. How much time does it take an electron to travel down the tube according to someone observing the proper tube length? c. How much time does the journey take from the electron’s point of view?
a. 37500; 299,792,457.9 m/s
The formula for the length contraction for a particle moving close to the speed of light is:
[tex]L' = \frac{L}{\gamma}[/tex]
where
L' is the length observed by the particle moving
L is the length observed by an observer at rest
In this problem, we have
[tex]L = 3 km = 3000 m[/tex] is the length of the SLAC measured by an observer at rest
[tex]L' = 8cm = 0.08 m[/tex] is the length measured by the electrons moving
Substituting into the formula, we find the gamma factor
[tex]\gamma = \frac{L}{L'}=\frac{3000 m}{0.08 m}=37,500[/tex]
The formula for the gamma factor is
[tex]\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}[/tex]
where v is the electron's speed and c is the speed of light. Re-arranging the equation, we find v:
[tex]1-\frac{v^2}{c^2}=\frac{1}{\gamma^2}\\v=c \sqrt{ 1-\frac{1}{\gamma^2}}=(299 792 458 m/s)\sqrt{1-\frac{1}{(37500)^2}}=299,792,457.9 m/s[/tex]
b. [tex]1.0\cdot 10^{-5}s[/tex]
For an observer at rest in the laboratory, the electron is moving at a speed of
v = 299,792,457.9 m/s
and it covers a total distance of
L = 3000 m
which is the length of the SLAC measured by the observer. Therefore, the time it takes for the electron to travel down the tube is
[tex]t=\frac{L}{v}=\frac{3000 m}{299,792,457.9 m/s}=1.0\cdot 10^{-5}s[/tex]
c. [tex]2.67\cdot 10^{-10}s[/tex]
From the electron's point of view, the length of the SLAC is actually contracted, so the electron "sees" a total distance to cover of
[tex]L' = 0.08 m[/tex]
And this means that the total time of travel of the electron, in its frame of reference will be shorter; in particular it is given by the formula:
[tex]t' = \frac{t}{\gamma}[/tex]
where
[tex]t=1.0\cdot 10^{-5}s[/tex] is the time measured by the observer at rest
[tex]\gamma=37500[/tex]
Substituting,
[tex]t' = \frac{1.0\cdot 10^{-5}s}{37500}=2.67\cdot 10^{-10}s[/tex]
A spring of constant k hangs from a ceiling with a mass m attached at the other end. The natural length of the hanging spring without the mass is L. a) Draw this scenario, including forces and known or calculable displacements. b) What is the equilibrium length of the spring with the mass attached? c) Suppose the spring is pulled down to create a displacement y from this equilibrium position and then released. Write an expression y (t) describing its position as a function of time t. Let "up" be the positive vertical direction. d) What is the period of oscillation for this simple harmonic oscillator? e) Suppose you cut the spring in half. What would be the spring constant of each half-spring? (The two are identical, so they will have the same spring constant.) To determine this, redraw your diagram as two springs of equal length connected "in series," i.e., top to bottom. Then indicate the forces acting on each of the three interesting bodies: the mass m and each halfspring of constant k’. Apply Hooke’s Law to each half-spring individually and then to the two springs together. f) Take the two half-springs from part e) and connect them "in parallel" (that is, side by side, hanging from the ceiling and each connected to the mass m). Draw the new diagram and repeat parts a) - d)
a) See part a of the attached picture.
There are two forces involved here:
- The weight of the block attached to the spring, of magnitude
W = mg
where m is the mass of the block and g = 9.8 m/s^2 is the acceleration due to gravity, pointing downward
- The restoring force of the spring,
F = kx
with k being the spring constant and x the displacement of the spring with respect to the equilibrium position, pointing upward
b) [tex]L' = L+\frac{mg}{k}[/tex]
The spring is in equilibrium when the restoring force of the spring is equal to the weigth of the block:
[tex]kx = mg[/tex]
where
x is the displacement of the spring with respect to the natural length of the spring, L. Solving for x,
[tex]x=\frac{mg}{k}[/tex]
And since the natural length is L, the equilibrium length of the spring is
[tex]L' = L+x=L+\frac{mg}{k}[/tex]
c) [tex]y(t) = y cos(\sqrt{\frac{k}{m}} t + \pi)[/tex]
Let's assume that
y = 0
corresponds to the equilibrium position of the spring (which is stretched by an amount [tex]L+\frac{mg}{k}[/tex]). If doing so, the vertical position of the mass at time t is given by
[tex]y(t) = y cos(\omega t + \pi)[/tex]
where
[tex]\omega = \sqrt{\frac{k}{m}}[/tex] is the angular frequency
t is the time
y is the initial displacement with respect to the equilibrium position
[tex]\pi[/tex] is the phase shift, so that the position at time t=0 is negative:
y(0) = -y
So rewriting the angular frequency:
[tex]y(t) = y cos(\sqrt{\frac{k}{m}} t + \pi)[/tex]
d) [tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex]
The period of oscillation is given by
[tex]T=\frac{2\pi}{\omega}[/tex]
where
[tex]\omega = \sqrt{\frac{k}{m}}[/tex] is the angular frequency
Substituting [tex]\omega[/tex], we find an expression for the period
[tex]T=2 \pi \sqrt{\frac{m}{k}}[/tex]
e) k' = 2k (see part e of attached picture)
Here the two half springs of spring constant k' are connected in series, so the sum of the stretchings of the two springs is equal to the total stretching of the spring, x:
[tex]x=x_1 + x_2[/tex]
Also, since the two springs are identical, their stretching will be the same:
[tex]x_1 = x_2 = x'[/tex]
so we have
[tex]x=x'+x'=2x'[/tex]
[tex]x'=\frac{x}{2}[/tex]
Substituting x find in part (b),
[tex]x'=\frac{mg}{2k}[/tex] (1)
Hooke's law for each spring can be written as
[tex]F=k' x'[/tex] (2)
where
F = mg (3)
is still the weight of the block
Using (1), (2) and (3) together, we find an expression for the spring constant k' of each spring
[tex]mg=k'\frac{mg}{2k}\\k' =2k[/tex]
So, the spring constant of each half-spring is twice the spring constant of the original spring.
fa) See part f) of attached picture
This time we have the block hanging from both the two half-springs, each with spring constant k'. So at equilibrium, the weight of the block is equal to the sum of the restoring forces of the two springs:
[tex]k' x_1 + k' x_2 = mg[/tex]
fb) [tex]x'=\frac{L}{2}+\frac{mg}{2k'}[/tex]
For the two springs in parallel, the sum of the restoring forces of the two springs must be equal to the weight of the block:
[tex]F_1 + F_2 = mg\\k_1 x_1 + k_2 x_2 = mg[/tex]
The two springs are identical, so they have same spring constant:
[tex]k_1 = k_2 = k'[/tex]
So (1) can be rewritten as
[tex]k' (x_1 + x_2) = mg[/tex]
And since the two springs are identical, their stretching x' is the same:
[tex]x_1 = x_2 = x'[/tex]
so we can rewrite this as
[tex]k' (2x') = mg[/tex]
[tex]x'=\frac{mg}{2k'}[/tex]
and so the equilibrium length of each spring will be
[tex]x'=\frac{L}{2}+\frac{mg}{2k'}[/tex]
fc) [tex]y(t) = y cos(\sqrt{\frac{2k'}{m}} t + \pi)[/tex]
The system of two springs in parallel can be treated as a system of a single spring with equivalent spring constant given by
[tex]k_{eq}=k_1 + k_2 = 2k'[/tex]
where k' is the spring constant of each spring.
So, let's assume again that
y = 0
corresponds to the equilibrium position as calculated in the previous part. If doing so, the vertical position of the mass at time t is given by:
[tex]y(t) = y cos(\omega t + \pi)[/tex]
where this time we have
[tex]\omega = \sqrt{\frac{k_{eq}}{m}}[/tex] is the angular frequency of the system
t is the time
y is the initial displacement with respect to the equilibrium position
[tex]\pi[/tex] is the phase shift, which we put so that the position at time t=0 is negative:
y(0) = -y
If we rewrite the angular frequency,
[tex]\omega = \sqrt{\frac{2k'}{m}}[/tex]
the position of the mass is
[tex]y(t) = y cos(\sqrt{\frac{2k'}{m}} t + \pi)[/tex]
fd) [tex]T=2 \pi \sqrt{\frac{m}{2k'}}[/tex]
Similarly to part d), the period of oscillation is
[tex]T=\frac{2\pi}{\omega}[/tex]
where
[tex]\omega = \sqrt{\frac{k_{eq}}{m}}=\sqrt{\frac{2k'}{m}}[/tex] is the angular frequency
Substituting [tex]\omega[/tex], we find
[tex]T=2 \pi \sqrt{\frac{m}{2k'}}[/tex]
Suppose a star the size of our Sun, but with mass 8.0 times as great, were rotating at a speed of 1.0 revolution every 8.3 days. If it were to undergo gravitational collapse to a neutron star of radius 15 km , losing 34 of its mass in the process, what would its rotation speed be? Assume the star is a uniform sphere at all times.
I don’t know what the answer is I Wish I could help
If a nichrome wire has a resistance of 15Ω and is connected across the terminals of a 3.0 V flashlight battery, how much current is in the wire?
1.0 amps
5.0 amps
7.5 amps
0.20 amps
Answer:
0.20 A
Explanation:
The current in the wire is given by Ohm's law:
[tex]I=\frac{V}{R}[/tex]
where
I is the current
V is the voltage
R is the resistance
In this problem,
V = 3.0 V is the voltage
[tex]R=15 \Omega[/tex] is the resistance
Substituting into the formula,
[tex]I=\frac{3.0 V}{15 \Omega}=0.20 A[/tex]
Final answer:
Using Ohm's Law, we calculate the current in a nichrome wire with 15Ω resistance connected to a 3.0 V battery to be 0.20 amps.
Explanation:
To determine the amount of current flowing through a nichrome wire when connected to a 3.0 V flashlight battery, we can use Ohm's Law, which states that the current (I) is equal to the voltage (V) divided by the resistance (R). Given that the resistance of the wire is 15Ω, we can calculate the current using the formula I = V/R.
Therefore, the current in the wire is as follows:
I = 3.0 V / 15Ω = 0.2 A
So, the current in the nichrome wire is 0.20 amps.
Calculate the heat flux (in W/m^2) through a sheet of a metal 14 mm thick if the temperatures at the two faces are 350 and 140°C. Assume steady-state heat flow and that the thermal conductivity of this metal is 52.4 W/m-K. (b) What is the heat loss per hour (in J/h) if the area of the sheet is 0.42 m2? (c) What will be the heat loss per hour (in J/h) if a material with a thermal conductivity of 1.8 W/m-K is used? (d) Calculate the heat loss per hour (in J/h) if the first metal is used and the thickness is increased to 24 mm.
Answer:
Explanation:
The rate of conductive heat transfer in watts is:
q = (k/s) A ΔT
where k is the heat conductivity, s is the thickness, A is the area, and ΔT is the temperature difference.
a)
Given k = 52.4 W/m/K, s = 0.014 m, and ΔT = 350-140 = 210 K, we can find q/A:
q/A = (52.4 / 0.014) (210)
q/A = 786,000 W/m²
b)
Given that A = 0.42 m², we can find q:
q = (0.42 m²) (786,000 W/m²)
q = 330,120 W
A watt is a Joule per second. Convert to Joules per hour:
q = 330,120 J/s * 3600 s/hr
q = 1.19×10⁹ J/hr
c)
If we change k to 1.8 W/m/K:
q = (k/s) A ΔT
q = (1.8 / 0.014) (0.42) (210)
q = 11,340 J/s
q = 4.08×10⁷ J/hr
d)
If k is 52.4 W/m/K and s is 0.024 m:
q = (k/s) A ΔT
q = (52.4 / 0.024) (0.42) (210)
q = 192,570 J/s
q = 6.93×10⁸ J/hr
The heat flux is 785,600 W/m^2, and the heat loss per hour is 1,110,912,000 J/h for the first metal. Using a material with a thermal conductivity of 1.8 W/m-K, the heat loss is 42,504,000 J/h. With increased thickness to 24 mm, the heat loss is 491,820,000 J/h.
Explanation:Firstly, you would calculate the heat flux through the metal sheet using Fourier's Law of Heat Conduction, which is expressed as:
q = -k * (T_2 - T_1) / d
In this case, the temperature difference (T_2 - T_1) is (350°C - 140°C), the thickness (d) of the metal sheet is 14 mm (which is 0.014 meters), and the thermal conductivity (k) is given as 52.4 W/m-K. Substituting these values in:
q = -52.4 * (350 - 140) / 0.014 = 785,600 W/m^2
For part (b), to calculate the heat loss per hour, we multiply the heat flux by the area of the sheet and convert seconds to hours:
Q/t = q * A = 785,600 * 0.42 m^2 * 3600 seconds = 1,110,912,000 J/h
For part (c), using a material with a thermal conductivity of 1.8 W/m-K, the process is similar, and the heat loss per hour would be:
Q/t = -1.8 * (350 - 140) / 0.014 * 0.42 * 3600 = 42,504,000 J/h
For part (d), with the increased thickness of 24 mm (0.024 meters), the heat loss per hour would be:
Q/t = -52.4 * (350 - 140) / 0.024 * 0.42 * 3600 = 491,820,000 J/h
A thick, spherical shell made of solid metal has an inner radius a = 0.18 m and an outer radius b = 0.33 m, and is initially uncharged. A point charge q = 5.00 C is placed at the center of the shell. What is the electric field strength in the region r < a? Express your answer in terms of 1/r2. Tries 0/8 What is the electric field strength in the region a < r < b? Express your answer in terms of 1/r2. Tries 0/8 What is the electric field strength in the region b < r? Express your answer in terms of 1/r2. Tries 0/8 What is the induced charge density at r = a? (in C/m^2) Tries 0/8 What is the induced charge density (in C/m2) at r = b? (in C/m^2)
A) [tex]E=\frac{4.50\cdot 10^{10}}{r^2} V/m[/tex]
r < a
We can find the magnitude of the electric field by using Gauss theorem. Taking a Gaussian spherical surface of radius r centered in the centre of the sphere, the electric flux through the surface of the sphere is equal to the ratio between the charge contained in the sphere and the vacuum permittivity:
[tex]E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}[/tex]
For r < a, the charge contained in the gaussian sphere is the point charge:
[tex]q=5.00 C[/tex]
So the electric field in this region is
[tex]E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{5.00 C}{4\pi (8.85\cdot 10^{-12} F/m)}\frac{1}{r^2}=\frac{4.50\cdot 10^{10}}{r^2} V/m[/tex]
B) E = 0
a < r < b
The region a < r < b is the region between the inner and the outer surface of the shell. We have to keep in mind that the presence of the single point charge +q = 5.00 C at the center of the sphere induces an opposite charge -q on the inner surface (r=a), and a charge of +q at the outer surface (r=b).
Using again Gauss theorem
[tex]E\cdot 4 \pi r^2 = \frac{q'}{\epsilon_0}[/tex]
this time we have that the gaussian sphere contains both the single point charge +q and the negative charge -q induced at r=a, so the net charge contained in the sphere is
q' = +q - q = 0
And so, the electric field in this region is zero.
C) [tex]E=\frac{4.50\cdot 10^{10}}{r^2} V/m[/tex]
r > b
Here we are outside of the sphere. Using Gauss theorem again
[tex]E\cdot 4 \pi r^2 = \frac{q'}{\epsilon_0}[/tex]
this time we have that the gaussian sphere contains the single point charge +q, the negative charge -q induced at r=a, and the positive charge +q induced at r=b, so the net charge contained in the sphere is
q' = +q - q +q = q
And so the electric field is identical to the one inside the sphere:
[tex]E=\frac{q}{4\pi \epsilon_0 r^2}=\frac{5.00 C}{4\pi (8.85\cdot 10^{-12} F/m)}\frac{1}{r^2}=\frac{4.50\cdot 10^{10}}{r^2} V/m[/tex]
D) -12.29 C/m^2
We said that the charge induced at the inner surface r=a is
-q = -5.00 C
The induced charge density is
[tex]\sigma = \frac{-q}{A}[/tex]
where A is the area of the inner surface of radius r = a = 0.18 m, so it is
[tex]A=4\pi a^2 = 4 \pi (0.18 m)^2=0.407 m^2[/tex]
So the induced charge density is
[tex]\sigma = \frac{-5.00 C}{0.407 m^2}=-12.29 C/m^2[/tex]
E) +3.65 C/m^2
We said that the charge induced at the outer surface r=b is
+q = +5.00 C
The induced charge density is
[tex]\sigma = \frac{+q}{A}[/tex]
where A is the area of the outer surface of radius r = b = 0.33 m, so it is
[tex]A=4\pi b^2 = 4 \pi (0.33 m)^2=1.368 m^2[/tex]
So the induced charge density is
[tex]\sigma = \frac{+5.00 C}{1.368 m^2}=+3.65 C/m^2[/tex]
A marathon runner completes a 42.238 km course in 2 h, 31 min, and 46 s . There is an uncertainty of 29 m in the distance run and an uncertainty of 3 s in the elapsed time.
What is the percent uncertainty in the average speed?
Answer:
The percentage uncertainty in the average speed is 0.10% (2 sig. fig.)
Explanation:
Consider the formula for average speed [tex]\bar{v}[/tex].
[tex]\displaystyle \bar{v} = \frac{s}{t}[/tex],
where
[tex]s[/tex] is the total distance, and [tex]t[/tex] is the time taken.The percentage uncertainty of a fraction is the sum of percentage uncertainties in
the numerator, andthe denominator.What are the percentage uncertainties in [tex]s[/tex] and [tex]t[/tex] in this question?
The unit of the absolute uncertainty in [tex]s[/tex] is meters. Thus, convert the unit of [tex]s[/tex] to meters:
[tex]s = \rm 42.238\;km = 42.238\times 10^{3}\;m[/tex].
[tex]\begin{aligned}\displaystyle \text{Percentage Uncertainty in }s &= \frac{\text{Absolute Uncertainty in } s}{\text{Measured Value of }s}\times 100\% \\ &=\rm\frac{29\; m}{42.238\times 10^{3}\;m}\times 100\%\\ &= 0.0687\%\end{aligned}[/tex].
The unit of the absolute uncertainty in [tex]t[/tex] is seconds. Convert the unit of [tex]t[/tex] to seconds:
[tex]t = \rm 2\times 3600 + 31\times 60 + 46 = 9106\;s[/tex]
Similarly,
[tex]\begin{aligned}\displaystyle \rm \text{Percentage Uncertainty in }t &= \frac{\text{Absolute Uncertainty in }t}{\text{Measured Value of }t}\times 100\% \\ &=\rm\frac{46\; s}{9106\;s}\times 100\%\\ &= 0.0329\%\end{aligned}[/tex].
The average speed [tex]\bar{v}[/tex] here is a fraction of [tex]s[/tex] and [tex]t[/tex]. Both [tex]s[/tex] and [tex]t[/tex] come with uncertainty. The percentage uncertainty in [tex]\bar{v}[/tex] will be the sum of percentage uncertainties in [tex]s[/tex] and [tex]t[/tex]. That is:
[tex]\text{Percentage Uncertainty in }\bar{v}\\=(\text{Percentage Uncertainty in } s) + (\text{Percentage Uncertainty in } t)\\ = 0.0687\% + 0.0329\%\\ = 0.010\%[/tex].
Generally, keep
two significant figures for percentage uncertainties that are less than 2%, and one for those that are greater than 2%.The percentage uncertainty in [tex]\bar{v}[/tex] here is less than 2%. Thus, keep two significant figures. However, keep more significant figures than that in calculations to make sure that the final result is accurate.
A 124-kg balloon carrying a 22-kg basket is descending with a constant downward velocity of 20.0 m/ s. A I.O-kg stone is thrown from the basket with an initial velocity of 15.0 m/ s perpendicular to the path of the descending balloon, as measured relative to a person at rest in the basket. The person in the basket sees the stone hit the ground 6.00 s after being thrown. Assume ,that the balloon continues its downward descent with the same constant speed of 20.0 m/ s. (a) How high was the balloon when the rock was thrown out? (b) How high is the balloon when the rock hits the ground') (c) At the instant the rock hits the ground, how far is it from the basket? (d) Just before the rock hits the ground, find its horizontal and vertical velocity components as measured by an observer (i) at rest in the basket and (ii) at rest on the ground
The balloon was 296.4 m high when the rock was thrown out; b) The balloon is 176.4 m high when the rock hits the ground; c) The distance between the rock and the basket when the rock hits the ground is 198.03 m; d) The horizontal velocity of the rock as measured by an observer at rest in the basket and ground is 15.0 m/s.
a) The initial velocity of the balloon is 20.0 m/s downward. The initial velocity of the rock is 15.0 m/s perpendicular to the path of the descending balloon. The person in the basket sees the stone hit the ground 6.00 s after being thrown.
Let the height of the balloon when the rock was thrown be h. The time it takes the rock to hit the ground is:
t = h / 15.0 m/s
Plugging in the value of h, we get:
t = 296.4 m / 15.0 m/s = 19.76 s
The total time it takes the rock to hit the ground is 6.00 s + 19.76 s = 25.76 s.
The vertical velocity of the rock is constant. The vertical velocity of the rock is:
v_y = 15.0 m/s
The horizontal velocity of the rock is also constant. The horizontal velocity of the rock is:
v_x = 15.0 m/s * (6.00 s / 25.76 s) = 3.53 m/s
The horizontal distance traveled by the rock is:
d_x = v_x * t
d_x = 3.53 m/s * 25.76 s = 90.36 m
The height of the balloon, when the rock was thrown, is:
h = d_x + v_y t
h = 90.36 m + 15.0 m/s * 25.76 s = 296.4 m
Therefore, the balloon was 296.4 m high when the rock was thrown out.
(b)The vertical velocity of the rock is constant. The vertical velocity of the rock is:
v_y = 15.0 m/s
The time it takes the rock to hit the ground is:
t = v_y / g
t = 15.0 m/s / 9.8 m/s² = 1.53 s
The height of the balloon when the rock hits the ground is:
h = v_y t
h = 15.0 m/s * 1.53 s = 22.95 m
Therefore, the balloon is 22.95 m high when the rock hits the ground.
(c) The horizontal velocity of the rock is constant. The horizontal velocity of the rock is:
v_x = 3.53 m/s
The time it takes the rock to hit the ground is:
t = 1.53 s
The horizontal distance traveled by the rock is:
d_x = v_x * t
d_x = 3.53 m/s * 1.53 s = 5.40 m
The distance between the rock and the basket when the rock hits the ground is:
d = d_x + h
d = 5.40 m + 22.95 m = 28.35 m
Therefore, the distance between the rock and the basket when the rock hits the ground is 28.35 m.
(d) i) The horizontal velocity of the rock as measured by an observer at rest in the basket is 15.0 m/s.
ii) The vertical velocity of the rock as measured by an observer at rest on the ground is 15.0 m/s.
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A water pipe tapers down from an initial radius of R1 = 0.21 m to a final radius of R2 = 0.11 m. The water flows at a velocity v1 = 0.84 m/s in the larger section of pipe. 1) What is the volume flow rate of the water?
Answer:
[tex]0.116 m^3/s[/tex]
Explanation:
The volume flow rate of a fluid in a pipe is given by:
[tex]Q=Av[/tex]
where
A is the cross-sectional area of the pipe
v is the speed of the fluid
In this problem, at the initial point we have
v = 0.84 m/s is the speed of the water
r = 0.21 m is the radius of the pipe, so the cross-sectional area is
[tex]A=\pi r^2 = \pi (0.21 m)^2 =0.138 m^2[/tex]
So, the volume flow rate is
[tex]Q=(0.138 m^2)(0.84 m/s)=0.116 m^3/s[/tex]
A dockworker loading crates on a ship finds that a 21-kg crate, initially at rest on a horizontal surface, requires a 73-N horizontal force to set it in motion. However, after the crate is in motion, a horizontal force of 55 N is required to keep it moving with a constant speed. Find the coefficients of static and kinetic friction between crate and floor. static friction kinetic friction
1) Static friction coefficient: 0.355
The crate is initially at rest. The crate remains at rest until the horizontal pushing force is less than the maximum static frictional force.
The maximum static frictional force is given by
[tex]F_s = \mu_s mg[/tex]
where
[tex]\mu_s[/tex] is the static coefficient of friction
m = 21 kg is the mass of the crate
g = 9.8 m/s^2 is the acceleration due to gravity
The horizontal force required to set the crate in motion is 73 N: this means that this is the value of the maximum static frictional force. So we have
[tex]F_s=73 N[/tex]
Using this information into the previous equation, we can find the coefficient of static friction:
[tex]\mu_s = \frac{F}{mg}=\frac{73 N}{(21 kg)(9.8 m/s^2)}=0.355[/tex]
2) Kinetic friction coefficient: 0.267
Now the crate is in motion: this means that the kinetic friction is acting on the crate, and its magnitude is
[tex]F_k = \mu_k mg[/tex] (1)
where
[tex]\mu_k[/tex] is the coefficient of kinetic friction
There is a horizontal force of
F = 55 N
pushing the crate. Moreover, the speed of the crate is constant: this means that the acceleration is zero, a = 0.
So we can write Newton's second law as
[tex]F-F_k = ma = 0[/tex]
And by substituting (1), we can find the value of the coefficient of kinetic friction:
[tex]F-\mu_k mg = 0\\\mu_k = \frac{F}{mg}=\frac{55 N}{(21 kg)(9.8 m/s^2)}=0.267[/tex]
The coefficients of static and kinetic friction between the crate and floor are;
μ_s = 0.3547
μ_k = 0.2672
1) To find the coefficient of static friction, is given by the formula;
F_s = μ_s*mg
We are given;
Mass; m = 21 kg
Horizontal force to set it to motion; F_h = 73 N
Now, in this static friction equation, we are using this force of 73 N because static friction is friction at rest and 73 N is the force to pull the object from rest. Thus;
73 = μ_s(21 × 9.8)
μ_s = 73/(21 × 9.8)
μ_s = 0.3547
2) To find the coefficient of kinetic friction, is given by the formula;
F_k = μ_k*mg
We will make use of the force of 55N because we are dealing with friction associated with motion and 55N is the force that keeps the object in motion. Thus;
μ_k = F_k/(mg)
μ_k = 55/(21 × 9.8)
μ_k = 0.2672
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If the mass of a material is 87 grams and the volume of the material is 14 cm3, what would the density of the material be?
Answer:
6.214g/cm³
Explanation:
The question is on density of a material
Density=mass/volume
Given, mass=87grams and volume= 14 cm³ density=?
Density=m/v 87/14 =6.214g/cm³
The radiation per unit area from the Sun reaching the earth is 1400 W/m2 , approximately the amount of radiative power per unit area reaching a sun bather on the banks of Barton Springs at noon on a clear day in June. The temperature of the Sun is 5800 K. Now suppose instead of the present Sun we received radiation from sun X at temperature 2864 K, located at the same position as our Sun. How much radiative power per unit area would reach the sun bather from the new sun X?
Answer:
83.2 W/m^2
Explanation:
The radiation per unit area of a star is directly proportional to the power emitted, which is given by Stefan-Boltzmann law:
[tex]P=\sigma A T^4[/tex]
where
[tex]\sigma[/tex] is the Stefan-Boltzmann constant
A is the surface area
T is the surface temperature
So, we see that the radiation per unit area is proportional to the fourth power of the temperature:
[tex]I \propto T^4[/tex]
So in our problem we can write:
[tex]I_1 : T_1^4 = I_2 : T_2^4[/tex]
where
[tex]I_1 = 1400 W/m^2[/tex] is the power per unit area of the present sun
[tex]T_1 = 5800 K[/tex] is the temperature of the sun
[tex]I_2[/tex] is the power per unit area of sun X
[tex]T_2 = 2864 K[/tex] is the temperature of sun X
Solving for I2, we find
[tex]I_2 = \frac{I_1 T_2^4}{T_1^4}=\frac{(1400 W/m^2)(2864 K)^4}{(5800 K)^4}=83.2 W/m^2[/tex]
An ice sled powered by a rocket engine starts from rest on a large frozen lake and accelerates at +35 ft/s2. After some time t1, the rocket engine is shut down and the sled moves with constant velocity v for a time t2. If the total distance traveled by the sled is 16350 ft and the total time is 90 s. Find the following. (a) Find the times t1 and t2. t1 t2 (b) Find the velocity v. ft/s At the 15750 ft mark, the sled begins to accelerate at -21 ft/s2. (c) What is the final position of the sled when it comes to rest? ft (d) What is the duration of the entire trip?
(a) 5.35 s, 84.65 s
The distance covered by the rocket during the first part of the motion (accelerated motion) is
[tex]d_1 = \frac{1}{2}at_1 ^2[/tex] (1)
where
a = +35 ft/s^2 is the acceleration
While the distance covered during the second part of the motion (uniform motion) is
[tex]d_2 = v t_2[/tex]
where v is the constant speed reached after the first part of the motion, which is given by
[tex]v=at_1[/tex]
So we can rewrite the equation as
[tex]d_2 = a t_1 t_2[/tex] (2)
We also know:
[tex]d= d_1 + d_2 = 16350[/tex] (3) is the total distance
[tex]t= t_1 +t_2 = 90[/tex] (4) is the total time
We can rewrite (3) as
[tex]\frac{1}{2}at_1^2 + a t_1 t_2 = 16350[/tex] (5)
And (4) as
[tex]t_2 = 90-t_1[/tex]
Substituting into (5), we get
[tex]\frac{1}{2}at_1^2 + a t_1 (90-t_1) = 16350\\0.5 at_1^2 + 90at_1 - at_1^2 = 16350\\-0.5at_1^2 +90at_1 = 16350[/tex]
Substituting a=+35 ft/s^2, we have
[tex]-17.5 t_1^2 +3150 t_1 - 16350 = 0[/tex]
which has two solutions:
[tex]t_1 = 174.65 s[/tex] --> discarded, since the total time cannot be greater than 90 s
[tex]t_1 = 5.35 s[/tex] --> this is the correct solution
And therefore [tex]t_2[/tex] is
[tex]t_2 = 90-t_1 = 90 -5.35 s=84.65 s[/tex]
(b) 187.25 ft/s
The velocity v is the velocity at the end of the accelerated motion of the rocket, so after
[tex]t_1 = 5.35 s[/tex]
The acceleration is
a = +35 ft/s^2
Therefore, the velocity after the first part is:
[tex]v=at_1 = (+35 ft/s^2)(5.35 s)=187.25 ft/s[/tex]
(c) 16,585 ft
At the 15750 ft mark, the velocity of the sled is
[tex]v=187.25 ft/s[/tex]
Then, it starts decelerating with acceleration
a = -21 ft/s^2
So, its distance covered during this part of the motion will be given by:
[tex]v_f ^ 2 -v^2 = 2ad_3[/tex]
where
vf = 0 is the final speed
v = 187.25 ft/s is the initial speed
a = -21 ft/s^2 is the acceleration
d3 is the distance covered during this part
Solving for d3,
[tex]d_3 = \frac{v_f^2 -v^2}{2a}=\frac{0-(187.25 ft/s)^2}{2(-21 ft/s^2)}=835 ft[/tex]
So, the final position of the sled will be
[tex]x_3 = 15750 ft + 835 ft =16,585 ft[/tex]
(d) 93.92 s
The duration of the first part of the motion is
[tex]t_1 = 5.35 s[/tex]
The distance covered during this part is
[tex]d_1 = \frac{1}{2}at_1^2=\frac{1}{2}(+35 ft/s^2)(5.35 s)^2=501 ft[/tex]
In the second part (uniform motion), the sled continues with constant speed until the 15750 ft mark, so the distance covered is
[tex]d_2 = 15750- 835 =14915[/tex]
And so the duration of the second part is
[tex]t_2 =\frac{d_2}{v}=\frac{14915}{187.25}=79.65 s[/tex]
While the duration of the third part (decelerated motion) is
[tex]a=\frac{v_f-v}{t_3}\\t_3 = \frac{v_f-f}{a}=\frac{0-(187.25 ft/s)}{-21 ft/s^2}=8.92 s[/tex]
So the total duration is
t = 5.35 s+79.65 s+8.92 s=93.92 s
This physics problem involving kinematics can be solved returning to the kinematic equations and applying them to the scenarios described. Acceleration, constant velocity and deceleration periods of motion each have specific times and these can be used to derive the specifics of the journey such as times, velocity and final position.
Explanation:The subject of this question is Physics, specifically, it is related to the concept of Kinematics which falls under Classical Mechanics.
In the first scenario, the sled accelerates from rest under the power of the rocket engine. We can use the kinematic equation: d = (1/2)at2 to solve for time t1, where a is the acceleration and d is the distance. The distance travelled under the acceleration is total distance minus the distance it goes with constant speed.
After finding t1, the time the sled moves with constant speed t2 is total time minus t1.
The constant speed v can be found by the distance traveled under constant speed divided by t2. When the sled hits the 15750ft mark, it starts to decelerate at -21 ft/s2. We can use the equation v2 = u2 + 2ad to get the distance it travels before it comes to rest.
Add this distance to 15750ft to get the final position of the sled when it comes to rest. Since we have the acceleration and final velocity is zero, one may get the time of this stage by using equation v = u + at . And the total duration is the sum of all time segments.
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Which of the following statements are true? Check all that apply. View Available Hint(s) Check all that apply. The average speed of gas molecules decreases with decreasing temperature. The kinetic energy of a molecule cannot determine its speed. All the gas molecules in a sample cannot have the same kinetic energy. The average kinetic energy of gas molecules decreases with decreasing temperature. There are gas molecules that move slower than the average.
The average speed and kinetic energy of gas molecules decrease with decreasing temperature, and there are gas molecules that move slower than the average.
Explanation:The correct statements are:
The average speed of gas molecules decreases with decreasing temperature. This is because at lower temperatures, the molecules have less kinetic energy and move slower.
The average kinetic energy of gas molecules decreases with decreasing temperature. As temperature decreases, the average kinetic energy of the gas molecules also decreases.
There are gas molecules that move slower than the average. In a sample of gas, there will always be molecules that have speeds below the average.
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Several statements about gas molecules and temperature are true, including the decrease in average speed and kinetic energy with decreasing temperature. Molecular speeds and kinetic energies vary, ensuring not all molecules have the same energy and some move slower than average. The kinetic energy of a molecule can determine its speed.
Let's analyze each statement one by one to determine which are true:
The average speed of gas molecules decreases with decreasing temperature. - This statement is true because as temperature decreases, the kinetic energy of the gas molecules also decreases, resulting in a lower average speed.The kinetic energy of a molecule cannot determine its speed. - This statement is false. The kinetic energy of a molecule is directly related to its speed by the equation KE = 1/2 mv², where m is the mass and v is the velocity.All the gas molecules in a sample cannot have the same kinetic energy. - This statement is true. At any given temperature, the molecules will have a distribution of kinetic energies.The average kinetic energy of gas molecules decreases with decreasing temperature. - This statement is true as the kinetic energy of gas molecules is directly proportional to temperature.There are gas molecules that move slower than the average. - This statement is true as molecular velocities are distributed over a range and there will always be molecules moving slower or faster than the average speed.The electric field strength between the plates of a simple air Capacitor is equal to the voltage across the plates divided by the distance between them. When a voltage of 112 V is put across the plates of such a capacitor an electric field strength of 1.1 kV/cm is measured. Write an equation that will let you calculate the distance d between the plates. Your equation should contain only symbols. Be sure you define each symbol
Answer:
[tex]d=\frac{V}{E}[/tex]
Explanation:
The electric field strength between the plates of a parallel plate capacitor is given by:
[tex]E=\frac{V}{d}[/tex]
where
V is the potential difference
d is the distance between the plates
If we re-arrange the formula, we find a new equation that lets us calculate d:
[tex]d=\frac{V}{E}[/tex]
We can at this point substitute the values:
V = 112 V
E = 1.1 kV/cm = 1100 V/cm = 110,000 V/m
To find the distance:
[tex]d=\frac{112 V}{110,000 V/m}=1.02\cdot 10^{-3} m[/tex]
The distance between capacitor plates, using the formula d = V/E, is approximately 1 cm
To calculate the distance d between the plates of a parallel plate capacitor, we use the relationship between the electric field strength E, the voltage V, and the distance d: E = V/d. Given:
V = 112 V (voltage across the plates)
E = 1.1 kV/cm = 1.1 * 10⁵ V/m (electric field strength)
The formula can be rearranged to solve for d:
d = V/E
Where:
V is the potential difference across the plates (in volts, V)E is the electric field strength (in volts per meter, V/m)d is the separation distance between the plates (in meters, m)Substituting the given values into the equation will allow us to calculate the distance between the plates.
Given:
Voltage (V) = 112 V
Electric field strength (E) = 1.1 kV/cm = 1.1 × 10⁴ V/m
Distance (d) = ?
Formula: E = V/d
Substitute: 1.1 × 10⁴ = 112/d
Solve for d: d = 112 / (1.1 × 10⁴)
[tex]\approx[/tex] 0.01 m [tex]\approx[/tex] 1 cm
The distance between the capacitor plates is thus approximately 1 cm.
which shows the correct order of steps when using wind to generate electricity
Answer:
kinetic energy from wind turns blades of wind turbines → wind turbine blades turn generator → generator changes kinetic energy to electricity
Answer:
kinetic energy from wind turns blades of wind turbines → wind turbine blades turn generator → generator changes kinetic energy to electricity
Explanation:
Wind energy is the energy harnessed from the velocity of wind. It is obtained with help of wind mills that have large fan type structure. The fan type structure has blades fixed on frame. When the high velocity wind blows, it cause the fan blades to rotate. The rotating fan causes the generator to turn. This rotating generator produces electricity. Now according to the explanation second option is the correct answer that is kinetic energy from wind turns blades of wind turbines → wind turbine blades turn generator → generator changes kinetic energy to electricity
can someone help me?!!!!!
Answer:
93 m
Explanation:
We need to find the resultant displacement.
Let's take north as positive y-direction and east as positive direction. We have
- Displacement 1 is 51 m to the north, so the two components are
[tex]x_1 =0\\y_1 = 51 m[/tex]
- Displacement 2 is 45 m [tex]60^{\circ}[/tex] north of east, so the two components are
[tex]x_1 = (45)cos 60^{\circ}=22.5 m\\y_1 = (45) sin 60^{\circ}=39.0 m[/tex]
So to find the resultant displacement we have to sum the components along each direction:
[tex]x=x_1 + x_2 = 0+22.5 m = 22.5 m\\y = y_1 + y_2 = 51 m +39.0 m = 90.0 m[/tex]
And the magnitude of the resultant displacement is
[tex]m=\sqrt{x^2+y^2}=\sqrt{(22.5 m)^2+(90.0 m)^2}=92.8 m \sim 93 m[/tex]
A spherical capacitor contains a charge of 3.10 nC when connected to a potential difference of 250 V. If its plates are separated by vacuum and the inner radius of the outer shell is 4.00 cma)Calculate the capacitance: found to be 1.24*10^-11 Fb)Find the radius of the inner spere: found to be .0294mc)Calculate the electric field just outside the surface of the inner sphere. T
(a) [tex]1.24\cdot 10^{-11} F[/tex]
The general formula for the capacitance of a capacitor is:
[tex]C=\frac{Q}{V}[/tex]
where
Q is the charge stored on the capacitor
V is the potential difference across the capacitor
In this problem, we have
[tex]Q=3.10 nC = 3.10\cdot 10^{-9} C[/tex] is the charge stored
V = 250 V is the potential difference
Substituting, we find
[tex]C=\frac{3.10\cdot 10^{-9}C}{250 V}=1.24\cdot 10^{-11} F[/tex]
(b) 0.0294 m
The capacitance of a spherical capacitor is given by
[tex]C=\frac{4\pi \epsilon_0}{\frac{1}{a}-\frac{1}{b}}[/tex]
where
a is the radius of the inner shell
b is the radius of the outer shell
Here we have
b = 4.00 cm = 0.04 m
and the capacitance is
[tex]C=1.24\cdot 10^{-11} F[/tex]
So we can re-arrange the equation to find a, the radius of the inner sphere:
[tex]a=(\frac{4\pi \epsilon_0}{C}+\frac{1}{b})^{-1} =(\frac{4\pi (8.85\cdot 10^{-12}F/m)}{1.24\cdot 10^{-11}F}+\frac{1}{0.04 m})^{-1}=0.0294 m[/tex]
(c) [tex]3.225 \cdot 10^4 V/m[/tex]
The electric field just outside the surface of the inner sphere with charge Q is equal to the electric field produced by a single point charge Q at a distance of r = a:
[tex]E=\frac{Q}{4\pi \epsilon_0 a^2}[/tex]
where
[tex]Q=3.10 \cdot 10^{-9} C[/tex] is the charge on the sphere
a = 0.0294 m is the radius of the inner sphere
Substituting the data into the formula, we find:
[tex]E=\frac{3.10\cdot 10^{-9}C}{4\pi (8.85\cdot 10^{-12}F/m) (0.0294 m)^2}=3.225 \cdot 10^4 V/m[/tex]