Answer
kinetic energy will be less than 1.2×10^4 J
Explanation:
as some of this potential energy will be used to over come force of friction , hence by law of conservation of energy , kinetic energy will be less than potential energy at top and will be less than 1.2×10^4 J
Final answer:
The ski jumper's kinetic energy at the bottom of the ski jump will be slightly less than the initial potential energy of 1.2 x 10⁴ J due to the small but non-negligible work done by friction, which converts some of the mechanical energy into other forms like heat and sound.
Explanation:
If a ski jumper has 1.2 x 10⁴ J of potential energy at the top of the ski jump and we take into account that friction is small but not negligible, we can conclude that the ski jumper's kinetic energy at the bottom will be slightly less than 1.2 x 10⁴ J. According to the conservation of energy principle, the sum of potential and kinetic energy in a system should be constant if there is no external work done. However, friction does perform negative work (it removes energy from the system), converting some mechanical energy into heat and sound, and therefore the actual kinetic energy at the bottom will be the initial potential energy minus the energy lost due to friction.
For example, in the given scenario where a ski jumper starts from rest, their initial kinetic energy is zero, and all the energy is in the form of potential energy. As they descend, potential energy is converted into kinetic energy. When friction is present, it will do negative work on the system, represented by a slight decrease in the total mechanical energy by the time the skier reaches the bottom. The kinetic energy of the ski jumper at the bottom would be the initial potential energy minus the work done by friction during the descent.
In a theoretical scenario without friction, the skier's kinetic energy at the bottom would equal their initial potential energy minus zero (since no work is done by friction), resulting in the skier having kinetic energy equal to 1.2 x 10⁴ J at the bottom.
When testing an PNP transistor with an ohmmeter, what are the high or low resistance values expected for a good transistor?
Answer:
0.45 V and 0.9 V.
Explanation:
To test a PNP transistor with an ohmmeter,
Plug the positive lead from the multimeter to the transistor EMITTER (E). Plug the negative meter into the transistor BASE (B). If you are using a PNP resistor you must watch OL that is over limit, the voltage decrease will indicate between 0.45V and 0.9V if you are measuring it.
Testing a PNP transistor with an ohmmeter should yield high resistance values when the base is negative relative to the emitter or collector, and low resistance values when the base is positive relative to the emitter or collector.
Explanation:When testing a PNP transistor with an ohmmeter, the high or low resistance values that are expected for a good transistor are as follows: When you measure between the base and the emitter or collector, you should see a high resistance value (in the range of megaohms) if the base is negative with respect to the emitter or collector. Conversely, you should see a low resistance value (in the range of a few hundred Ohms) when the base is positive with respect to the emitter or collector.
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A motorcycle rides on the vertical walls around the perimeter of a large circular room. The friction coefficient between the motorcycle tires and the walls is µ. How does the minimum µ needed to prevent the motorcycle from slipping downwards change with the motorcycle’s speed, s?
a) µ ∝ s0b) µ ∝ s−1/2c) µ ∝ s−1d) µ ∝ s−2e) none of these
Answer:
option D
Explanation:
given,
coefficient of friction between wall and tire = µ
speed of motorcycle = s
friction force = f = μ N
where normal force will be equal to centripetal force
[tex]N = \dfrac{mv^2}{r}[/tex]
for motorcycle to not to slip weight should equal to the centripetal force
now,
[tex]m g =\mu \dfrac{mv^2}{r}[/tex]
[tex]\mu =\dfrac{rg}{s^2}[/tex]
where "rg" is constant
[tex]\mu\ \alpha \ \dfrac{1}{s^2}[/tex]
[tex]\mu\ \alpha \ s^{-2}[/tex]
Hence, the correct answer is option D
Two technicians are discussing oil leaks. Technician A says that an oil leak can be found using a fluorescent dye in the oil with a black light to check for leaks. Technician B says a white spray powder can be used to locate oil leaks. Which technician is correct?a. Technician A only
b. Technician B only
c. Both Technicians A and B
d. Neither Technician A nor B
Answer:
B
Explanation:
cause it makes the most sence
A space station in the form of a large wheel, 283 m in diameter, rotates to provide an "artif icial gravity" of 9.5 m/s 2 for people located at the outer rim. What is the frequency of the rotational motion for the wheel to produce this effect? Answer in units of rev/min.
Answer:
The frequency of the rotational motion for the wheel to produce this effect is 2.473 rev/min.
Explanation:
Given that,
Acceleration = 9.5 m/s²
Diameter = 283 m
We need to calculate the frequency of the rotational motion for the wheel to produce this effect
Using formula of rotational frequency
[tex]a= r\omega^2[/tex]
[tex]\omega=\sqrt{\dfrac{a}{r}}[/tex]
Where, r = radius
a = acceleration
[tex]\omega[/tex] = rotational frequency
Put the value into the formula
[tex]\omega=\sqrt{\dfrac{9.5\times2}{283}}[/tex]
[tex]\omega=0.259\ rad/s[/tex]
The frequency in rev/min
[tex]\omega=0.259\times\dfrac{60}{2\pi}[/tex]
[tex]\omega=2.473\ rev/min[/tex]
Hence, The frequency of the rotational motion for the wheel to produce this effect is 2.473 rev/min.
A stationary source S generates circular outgoing waves on a lake. The wave speed is 5.0 m/s and the crest-to-crest distance is 2.0 m. A person in a motorboat heads directly toward S at 3.0 m/s. To this person, the frequency of these waves is:
Answer:4 Hz
Explanation:
Speed of wave [tex]v=5 m/s[/tex]
crest to crest distance [tex]\lambda =2 m[/tex]
velocity of observer [tex]v_0=3 m/s[/tex]
actual frequency [tex]f=\frac{velocity}{\lambda }[/tex]
[tex]f=\frac{5}{2}=2.5 Hz[/tex]
Apparent frequency [tex]f'=f(\frac{v+v_0}{v})[/tex]
[tex]f'=2.5\times \frac{5+3}{5}[/tex]
[tex]f'=4 Hz[/tex]
The frequency of these waves is 4hz.
Given information:The wave speed is 5.0 m/s and the crest-to-crest distance is 2.0 m. Alo, A person in a motorboat head directly toward S at 3.0 m/s.
Calculation of wave frequency:The actual frequency is [tex]= 5 \div 2[/tex] = 2.5Hz
Now
Apparent frequency is
[tex]= 2.5 \times (5 + 3) \div 5[/tex]
= 4hz
Hence, the frequency of these waves is 4hz.
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The edge of a cube was found to be 30 cm with a possible error in measurement of 0.5 cm. Use differentials to estimate the maximum possible error, relative error, and percentage error in computing the volume of the cube and the surface area of the cube. (Round your answers to four decimal places.)
Answer:
In computing the volume of a cube,
Maximum possible error = +/-1350cm³
Relative error = 0.05
Percentage error = 5%
In computing the surface area of a cube,
Maximum possible error = +/-180cm²
Relative error = 0.0333
Percentage error = 3.33%
Explanation:
A cube is a three dimensional solid object with six (6) faces, twelve (12) edges and eight(8) vertices.
The volume of a cube = x³
Where x= length of the edge of a cube
X = 30cm +/- 0.5cm
Differentiate V with respect to x (V = Volume of a cube)
dV/dx = 3 x²
dV = 3 x² . dx
dV= 3 × 30² × (+/-0.5)
= 2700(+/-0.5)
= +/-1350cm³
Maximum possible error =
+/- 1350cm³
Relative error = Maximum error /surface area
= ΔV/V
Recall that V = x³
V= (30)³
A = 27000cm³
Substitute the values for and V into the formula for Relative error
Relative error = 1350 / 270000
Relative error = 0.05
% error = Relative error × 100
= 0.05× 100
= 5%
Surface Area of a cube = 6x²
A = 6x²
Differentiate A with respect to x
dA/dx= 12x
dA = 12x . dx
dA= 12 × 30 (0.5)
= +/- 180cm²
Maximum possible error =
+/- 180cm²
Relative error = Maximum error / total area
= dA/dx
Recall that A = 6x²
A = 6(30)²
A = 5400cm²
Substitute the values for and A into the formula for Relative error
Relative error = 180/ 5400
Relative error = 0.0333(4 decimal place)
% error = Relative error × 100
= 0.0333 × 100
= 3.33%
Final answer:
Use of differentials to estimate maximum and relative errors in volume and surface area calculations for a cube.
Explanation:
Differentials for Cube:
Maximum possible error in volume: 30*(0.5) = 15 cm³Relative error in volume: 15/30 = 0.5Percentage error in volume: (0.5)*100% = 50%Surface Area:
Maximum possible error in surface area: 6*(30)*(0.5) = 90 cm²Relative error in surface area: 90/(6*30) = 0.25Percentage error in surface area: (0.25)*100% = 25%A child pushes horizontally on a box of mass m which moves with constant speed v across a horizontal floor. The coefficient of friction between the box and the floor is μ. At what rate does the child do work on the box?
Answer:
Rate of child doing work on box = μmgv (Unit is Watt)
Explanation:
Rate of child doing work on box = Work done / time = Power
Power = Horizontal force x Velocity
We are aware that the Velocity in this case is v.
As the object is moving with constant velocity, the acceleration would be zero and the applied horizontal force will be equal to friction force. So in our case,
Horizontal force = friction force
We know that the coefficient of friction is the ratio of friction force to Normal force,
μ = friction force / Normal force
Normal Force = mg, where m is the mass and g is the gravitational acceleration
Friction force = μ x Normal Force
Friction force = μmg
Power = μmgv (Unit of power is Watt)
Answer:
P = μ*mg*v
Explanation:
A child pushes horizontally on a box of mass m which moves with constant speed v across a horizontal floor. The coefficient of friction between the box and the floor is μ. The rate at which the child works is calculated as shown below:
mass of the box = m; coefficient of friction is μ; speed = v.
In order to push the box, the child must exert a force equal to or more than the frictional force.
force = coefficient of friction*weight of the box
f = μ*mg
In addition, to calculate the rate of work (i.e. power). We have:
Power = force*velocity (or speed)
Therefore:
P = μ*mg*v
A 0.47 kg mass is attached to a spring with a spring constant of 130 N/m so that the mass is allowed to move on a horizontal friction-less surface. The mass is released from rest when the spring is compressed 0.12 m. A) Find the force on the mass at the instant the spring is released. B) Find the acceleration of the mass at the instant the spring is released.
Answer:
F= 15.6 N, a= 33.2 m/s^2
Explanation:
mass= m = 0.47 kg
spring constant= k = 130N/m
spring compression = x = 0.12 m
a).
force on the mass= F = k*x
F = 130 * 0.12 N
F= 15.6 N
b).
Acceleration of mass= a=?
F= ma
a=F / m
a= 15.6/ 0.47 m/s^2
a= 33.2 m/s^2
Final answer:
The force on the mass when the spring is released is 15.6 N, and the acceleration of the mass at that instant is approximately 33.19 m/s^2, calculated using Hooke's Law and Newton's second law, respectively.
Explanation:
Understanding Spring Force and Acceleration
To find the force on the mass at the instant the spring is released (in part A), we use Hooke's Law, which states that the force exerted by a spring (F) is equal to the negative spring constant (k) times the displacement from equilibrium (x), so F = -kx. Here, k = 130 N/m and x = 0.12 m, so the force is F = 130 N/m * 0.12 m = 15.6 N. The negative sign indicates that the force is in the opposite direction of the displacement.
To find the acceleration of the mass at the instant the spring is released (in part B), we apply Newton's second law, which relates force (F), mass (m), and acceleration (a) as F = ma. Rearranging for acceleration, we get a = F/m. Substituting the values, we have a = 15.6 N / 0.47 kg = approximately 33.19 m/s2.
Suppose that you have a reflection diffraction grating with n= 125 lines per millimeter. Light from a sodium lamp passes through the grating and is diffracted onto a distant screen. PART A
Two visible lines in the sodium spectrum have wavelengths 498\rm nm and 569 \rm nm. What is the angular separation \Delta \theta of the first maxima of these spectral linesgenerated by this diffraction grating?
PART B
How wide does this grating need to be to allow you to resolvethe two lines 589.00 and 589.59 nanometers, which are a well knownpair of lines for sodium, in the second order (m=2)?
Answer:
[tex]0.54^{\circ}[/tex]
3.99322032 mm
Explanation:
n = Lines per mm = 125
Seperation between slits is given by
[tex]d=\dfrac{1}{n}\\\Rightarrow d=\dfrac{1}{125}\\\Rightarrow d=0.008\ mm[/tex]
[tex]\lambda_1[/tex] = 498 nm
[tex]\lambda_2[/tex] = 569 nm
We have the expression
[tex]dsin\theta_1=m\lambda_1[/tex]
For first maximum m = 1
[tex]\theta_1=sin^{-1}\dfrac{m\lambda_1}{d}\\\Rightarrow \theta_1=sin^{-1}\dfrac{1\times 498\times 10^{-9}}{0.008\times 10^{-3}}\\\Rightarrow \theta_1=3.57^{\circ}[/tex]
[tex]\theta_2=sin^{-1}\dfrac{m\lambda_2}{d}\\\Rightarrow \theta_2=sin^{-1}\dfrac{1\times 569\times 10^{-9}}{0.008\times 10^{-3}}\\\Rightarrow \theta_2=4.08^{\circ}[/tex]
Angular separation is given by
[tex]\Delta \theta=\theta_2-\theta_1\\\Rightarrow \Delta \theta=4.08-3.57\\\Rightarrow \Delta \theta=0.54^{\circ}[/tex]
Angular separation is [tex]0.54^{\circ}[/tex]
Now
[tex]\lambda_1[/tex] = 589 nm
[tex]\lambda_2[/tex] = 589.59 nm
[tex]\Delta \lambda=\lambda_2-\lambda_1\\\Rightarrow \Delta \lambda=589.59-589\\\Rightarrow \Delta \lambda=0.59]\ nm[/tex]
We have the relation
[tex]\dfrac{\lambda}{\Delta \lambda}=mN\\\Rightarrow N=\dfrac{\lambda}{m\Delta \lambda}\\\Rightarrow N=\dfrac{589}{2\times 0.59}\\\Rightarrow N=499.15254[/tex]
Width is given by
[tex]w=\dfrac{N}{n}\\\Rightarrow w=\dfrac{499.15254}{125}\\\Rightarrow w=3.99322032\ mm[/tex]
The width is 3.99322032 mm
The angular separation \theta of the first maxima of these spectral lines generated by this diffraction grating is 0.54°
The width which this grating needs to be to allow you to resolve the two lines 589.00 and 589.59 nanometers is 3.99322032 mm
Calculations and Parameters:n = Lines per mm
= 125
The Separation between slits is given by:
d= 1/n
d= 1/125
= 0.008mm.
Where
line 1 = 498nm
line 2 = 569nm
The first maximum m= 1 will be:
θ1= 3.57°
θ2= 4.08°
The angular separation would be:
θ2- θ1= 0.54°.
Now, to find the width is:
w= N/n
= 499.15254/125
= 3.99322032 mm.
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the decimal reduction time (DRT) is the time it takes to kill 90% of cells present. Assume that a DRT value for autoclaving a culture is 1.5 minutes. How long would it take to kill all the cells if 10^6 cells were present? What would happen if you stopped the heating process at 9 minutes?
Answer:
It takes 10.5 minutes to kill all the bacteria.
Only 1 cell would remain after 9 minutes.
Explanation:
It will take 1.5 minutes to kill 90% of the cells. So, after 1.5 minutes, only 10% would remain. After 3 minutes, only 1% remain. So, to figure out how long it would take to kill a million cells, we have to multiply 1 million by 0.1 repeatedly until the final value is less than 1 that is because when the value is less than 1, it means there are no more bacteria.
So:
[tex]10^6 \times (0.1)^7[/tex] = 0.1
So, you need 10.5 minutes of killing to kill one million cells.
Time taken= 7 x 1.5 minutes = 10.5 minutes.
After 9 minutes you would have:
[tex]10^{6} \times (0.1)^{6}[/tex] = 1 cell left
The decimal reduction time (DRT) is the time it takes to kill 90% of cells present. If a DRT value for autoclaving a culture is 1.5 minutes, it means that it takes 1.5 minutes to kill 90% of the cells. To calculate how long it would take to kill all the cells if 10^6 cells were present, you can use the DRT value as a basis. If you stopped the heating process at 9 minutes, it means that you haven't reached the time required to kill 100% of the cells.
Explanation:The decimal reduction time (DRT) is the time it takes to kill 90% of cells present. If a DRT value for autoclaving a culture is 1.5 minutes, it means that it takes 1.5 minutes to kill 90% of the cells. To calculate how long it would take to kill all the cells if 10^6 cells were present, you can use the DRT value as a basis. Since the DRT value represents the time it takes to kill 90% of the cells, you can calculate the time to kill 100% of the cells by dividing the DRT value by 90 and then multiplying it by 100. In this case, it would be as follows:
T = (1.5 minutes / 90) * 100 = 1.67 minutes
If you stopped the heating process at 9 minutes, it means that you haven't reached the time required to kill 100% of the cells. As a result, some cells would still be alive.
The mass of a spacecraft is about 435 kg . An engine designed to increase the speed of the spacecraft while in outer space provides 0.09-N thrust at maximum power.By how much does the engine cause the craft's speed to change in 1 week of running at maximum power? Describe any assumptions you made.
Answer:
Δ v = 125 m/s
Explanation:
given,
mass of space craft = 435 Kg
thrust = 0.09 N
time = 1 week
= 7 x 24 x 60 x 60 s
change in speed of craft = ?
Assuming no external force is exerted on the space craft
now,
[tex]T= m_s a[/tex]
[tex]a=\dfrac{T}{m_s}[/tex]
[tex]a =\dfrac{0.09}{435}[/tex]
a = 2.068 x 10⁻⁴ m/s²
using equation of motion
Δ v = a t
Δ v = 2.068 x 10⁻⁴ x 7 x 24 x 60 x 60
Δ v = 125 m/s
Your car has stalled and you need to push it. You notice as the car gets going that you need less and less force to keep it going. Suppose that for the first 15 m your force decreased at a constant rate from 210 N to 45 N. How much work did you do on the car?
Answer:
675 Joules
Explanation:
Considering that work can be calculated with the following formula:
W = Fx D
Where:
W = work
F = force
D = distance
We can directly use this formula in case the applied force remains constant
In case the force does not remain constant, we can calculate the work as a change, this is :
ΔW = ΔFxΔD
For this scenario, we have:
W₁ = F₁xD₁
F₁ = 210 N, D = 0 m
W₁ = 210Nx0m = 0 Joules
W₂ = F₂xD₂
F₂ = 45 N , D₂ = 15 m
W₂ = 45Nx15m = 675 Joules
Finally: ΔW = Total work performed when moving the car 15 m
ΔW = W₂ - W₁ = 675 Joules - 0 Joules = 675 Joules
Final answer:
The amount of work done on the car while pushing it is 1575 Joules.
Explanation:
Work is defined as the product of force and displacement. In this case, the force required to keep the car moving decreases as the car gets going. Work can be calculated using the formula:
Work = Force × Distance
Given that the force decreased from 210 N to 45 N over a distance of 15 m, we can calculate the work done as follows:
Work = (210 N + 45 N) / 2 × 15 m = 1575 J
Therefore, the amount of work done on the car while pushing it is 1575 Joules.
Calculate the electric potential V(h) inside the capacitor as a function of height h. Take the potential at the bottom plate to be zero.Express V(h) in terms of E and h.V(h) =
Answer:
V(h) = Eh
Explanation:
I will assume that the capacitor is a parallel-plate capacitor.
By Gauss' Law, electric field inside the capacitor is
[tex]E = \frac{\sigma}{\epsilon_0} = \frac{Q}{\epsilon_0 A}[/tex]
The relation between electric field and potential is
[tex]V_{ab} = -\int\limits^b_a {\vec{E}(h)} \, d\vec{h} = \int\limits^h_0 {\frac{Q}{\epsilon_0 A}} \, dh \\V(h) - V(0) = V(h) - 0 = Eh\\V(h) = Eh[/tex]
The important thing in this question is that the electric field inside the parallel plate is constant. So, the potential is also constant and proportional to the distance, h.
The electric potential V(h) inside a capacitor as a function of height h, with zero potential at the bottom plate, is V(h) = Eh, using the relationship between the electric field E and potential V where E is constant.
Explanation:To calculate the electric potential V(h) inside a capacitor as a function of height h, with the potential at the bottom plate taken to be zero, you can use a relation between the electric field E and the potential V. Given that E = V/d, where d is the separation between the plates, and that the electric field E is uniform, we have the relationship E = -dV/dh (the negative sign indicates the direction of the potential decrease). Integrating this equation from 0 to h, where V(0) = 0, gives us V(h) = -Eh. However, we can ignore the negative sign because we are interested in magnitude.
So, the final expression for V(h) inside the capacitor, in terms of the electric field E and the height h above the bottom plate, is:
V(h) = Eh
The rumble feature on a video game controller is driven by a device that turns electrical energy into mechanical energy. This device is best referred to as
Answer:
MOTOR
Explanation:
The device which changes electrical energy into mechanical energy is known as motor
Loops of wire in a magnetic field make up motors. The magnetic field produces a torque on the loops when current flows through them, turning a shaft as a result.
What is electric motor?Electrical energy is transformed into mechanical energy by electric motors. When we turn on the fan, for instance, the electric motor begins to transform the electrical energy into mechanical energy.
The fan blades then begin whirling as a result of the mechanical energy, giving them the capacity to perform work.
On either side of the controller, there is a motor. This engine has an uneven weight linked to it.
This indicates that one side of it is heavier than the other. The imbalance of the weight causes the controller to vibrate when the motor turns.
Therefore, Mechanical work is created as a result of the conversion of electrical energy.
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N2 is non-reactive, however, if you can get it to react with H2, it does so in a 1:3 mole ratio (N2:H2) and releases 92.0 kJ of thermal energy. That means which if these statements are true?
A. This reaction is exothermic.
B. This reaction has a negative enthalpy of reaction.
C. All of the above are true.
D. This reaction has a negative enthalpy of reaction.
Answer:
C. all above is true.
Explanation:
Energy releasing reactions are exothermic. Total energy of products ( [tex] N_2 H_2[/tex] ) is less than the total energy of reactants ( [tex] H_2 + N_2 [/tex] ) gives negative enthalpy change.
hint: prefix exo- means "outside, external".
Approximate the work required to lift a 2.5-kg object to a height of 6.0 meters. A student applies a force to a cart to pull it up an inclined plane at a constant speed during a physics lab. A force of 20.8 N is applied parallel to the incline to lift a 3.00-kg loaded cart to a height of 0.450 m along an incline which is 0.636-m long. Determine the work done upon the cart and the subsequent potential energy change of the cart. Eddy, whose mass is 65-kg, climbs up the 1.6-meter high stairs in 1.2 s. Approximate Eddy's power rating.
Answer:
Explanation:
(a)mass [tex]m= 2.5 kg [/tex]
height [tex]h=6 m[/tex]
work required to raise
[tex]W=mgh[/tex]
[tex]W=2.5\times 9.8\times 6=147 J[/tex]
(b)Force [tex]F=20.8 N[/tex]
mass of cart [tex]m=3 kg[/tex]
length of track [tex]s=0.636 m[/tex]
[tex]Work\ done=F\cdot s[/tex]
Work done[tex]=20.8\cdot 0.636=13.22 J[/tex]
(c)mass of eddy [tex]m_e=65 kg[/tex]
height climbed [tex]h=1.6 m[/tex]
time [tex]t=1.2 s[/tex]
Energy required [tex]E=mgh=65\times 9.8\times 1.6=1019.2 J[/tex]
[tex]power=\frac{E}{t}=\frac{1019.2}{1.2}=849.33 W[/tex]
The work done in lifting the 2.5 kg object is 147 J, in moving the 3-kg cart is 13.2 J, and the power used by Eddy in climbing stairs is 851.7 W.
Explanation:To solve these problems, we need to apply principles of physics, specifically related to work, energy, and power. For the first question, we use the concept of gravitational potential energy, which is calculated by multiplying together the object's mass, the acceleration due to gravity (~9.81 m/s² on Earth), and the height to which the object is lifted. Thus for the 2.5-kg object, the work done or energy required to lift it to a height of 6.0 meters is: W = m * g * h = 2.5 kg * 9.8 m/s² * 6.0 m = 147 J.
Next, for the 3-kg cart, since the cart moves at constant speed, we can say the work done is equal to the change in potential energy. Therefore, the work done is W = m * g * h = 3.0 kg * 9.8 m/s² * 0.450 m = 13.2 J.
Fianlly, for Eddy's case, power is defined as the work done per unit time. If Eddy lifts his own mass to the height of 1.6 m, the work done (again considering as change in gravitational potential energy) is W = m * g * h = 65 kg * 9.8 m/s² * 1.6 m = 1022 J. Given that he does this work in 1.2 seconds, the power expended would be P = W / t = 1022 J / 1.2 s = 851.7 W.
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A man goes for a walk, starting from the origin of an xyz coordinate system, with the xy plane horizontal and the x axis eastward. Carrying a bad penny, he walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high
(a) In unit-vector notation, what is the displacement of the penny from start to its landing point?
(b) When the man returns to the origin, what is the magnitude of his displacement for the return trip?
Explanation:
a) He walks 1300 m east, 2400 m north, and then drops the penny from a cliff 640 m high.
1300 m east = 1300 i
2400 m north = 2400 j
Drops the penny from a cliff 640 m high = -640 k
Displacement of penny = 1300 i + 2400 j - 640 k
b) Displacement of man for return trip = -1300 i - 2400 j
[tex]\texttt{Magnitude = }\sqrt{(-1300)^2+(-2400)^2}=2729.47m[/tex]
Magnitude of his displacement = 2729.47 m
Answer:
Explanation:
d1 = 1300 m east
d2 = 2400 m north
d3 = 640 m downward
(a)
The displacement of penny is given by
[tex]\overrightarrow{d}=\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}[/tex]
[tex]\overrightarrow{d}=1300\widehat{i}+2400 \widehat{j}-640\widehat{k}[/tex]
(b) For the return journey of man, the displacement is given by
[tex]\overrightarrow{d}=-\overrightarrow{d_{1}}-\overrightarrow{d_{2}}[/tex]
[tex]\overrightarrow{d}=-1300\widehat{i}-2400 \widehat{j}[/tex]
The magnitude of the displacement is given by
[tex]d=\sqrt{1300^{2}+2400^{2}}=2729.47 m[/tex]
The force that attracts earth to an object is equal to and opposite the force that earth exerts on the object. Explain why earth's acceleration is not equal to and opposite the object's acceleration.
Answer:
Because of heavy mass
Explanation:
When force acts on a body it tends to accelerate the body. The acceleration produced in the body depends on two things:
1). Magnitude of force
2). Mass of the body
F= ma
⇒ a = F/m
As the force exerted on earth and another object are the equal in magnitude but opposite in direction. This forces will accelerate the object toward the earth but can't accelerate the earth as earth has very high mass.
a = F/m
This force tends to accelerate the earth but but due to earth's inertia the earth does not accelerate.
The temperature inside a vehicle can rise ____ degrees higher than the outside temperature.
Answer:
Generally, 40 to 50 degrees
Explanation:
About the heat-up over time, whether the windows of a vehicle are locked or partially open makes very little difference. In both situations, in an internal temperature of a vehicle, even at an outside temperature of only 72 ° F, it may exceed approximately 40 ° F within one hour. This happens mainly due to the greenhouse effect that is the heat inside the car is trapped and not allowed to escape. Thus, raising the temperature of the vehicle.
Final answer:
The greenhouse effect causes the temperature inside a parked car to be much higher than outside, sometimes 20 to 30 degrees higher, due to sunlight being trapped as heat.
Explanation:
The temperature inside a vehicle can rise significantly higher than the outside temperature due to a phenomenon known as the greenhouse effect. When a car is parked in the sun with windows closed, the sunlight passes through the glass and is absorbed by interior surfaces such as dashboard and seats. These surfaces then emit infrared radiation, which cannot escape back through the glass, and consequently, the trapped heat raises the temperature of the air inside the car. This effect can cause interior temperatures to be much higher—often 20 to 30 degrees higher—than the external air temperature.
Two point charges are separated by a distance of 10.0 cm. One has a charge of -25 μC and the other +50 μC. (a) Determine the direction and magnitude of the electric field at a point P between the two charges that is 2.0 cm from the negative charge. (b) If an electron (mass = 9.11 x 10-31 kg) is placed at rest at P and then released, what will be its initial acceleration (direction and magnitude)?
Answer:
a)
6.33 x 10⁸ N/C
Direction : Towards negative charge.
b)
1.11125 x 10²⁰ m/s²
Direction : Towards positive charge.
Explanation:
a)
[tex]Q_{1}[/tex] = magnitude of negative charge = 25 x 10⁻⁶ C
[tex]Q_{2}[/tex] = magnitude of positive charge = 50 x 10⁻⁶ C
[tex]r_{1}[/tex] = distance of negative charge from point P = 0.02 m
[tex]r_{2}[/tex] = distance of positive charge from point P = 0.08 m
Magnitude of electric field at P due to negative charge is given as
[tex]E_{1} = \frac{kQ_{1}}{r_{1}^{2} } = \frac{(9\times10^{9})(25\times10^{-6})}{0.02^{2} } = 5.625\times10^{8} N/C[/tex]
Magnitude of electric field at P due to positive charge is given as
[tex]E_{2} = \frac{kQ_{2}}{r_{2}^{2} } = \frac{(9\times10^{9})(50\times10^{-6})}{0.08^{2} } = 0.703125\times10^{8} N/C[/tex]
Net electric field at P is given as
[tex]E = E_{1} + E_{2}\\E = 5.625\times10^{8} + 0.703125\times10^{8} \\E = 6.33\times10^{8} N/C[/tex]
Direction:
Towards the negative charge.
b)
[tex]m[/tex] = mass of the electron placed at P = 9.31 x 10⁻³¹ C
[tex]Q_{1}[/tex] = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
Acceleration of the electron due to the electric field at P is given as
[tex]a = \frac{qE}{m}\\ a = \frac{(1.6\times10^{-19})(6.33\times10^{8})}{(9.11\times10^{-31})}\\a = 1.11125\times10^{20} ms^{-2}[/tex]
Direction: Towards the positive charge Since a negative charge experience electric force in opposite direction of the electric field.
Answer:
Explanation:
qA = - 25 x 10^-6 C
qB = 50 x 10^-6 C
AP = 2 cm
BP = 8 cm
(a)
Electric field at P due to the charge at A
[tex]E_{A}=\frac{kq_{A}}{AP^{2}}[/tex]
[tex]E_{A}=\frac{9\times 10^{9}\times 25\times 10^{-6}}{0.02^{2}}[/tex]
EA = 5.625 x 10^8 N/C
Electric field at P due to the charge at B
[tex]E_{B}=\frac{kq_{B}}{BP^{2}}[/tex]
[tex]E_{A}=\frac{9\times 10^{9}\times 50\times 10^{-6}}{0.08^{2}}[/tex]
EB = 0.70 x 10^8 N/C
The resultant electric field at P due to both the charges is
E = EA+ EB = (5.625 + 0.7) x 10^8
E = 6.325 x 10^8 N/C towards left
(b) mass of electron, m = 9.1 x 10^-31 kg
Let a be the acceleration of electron
Force on electron, F = charge of electron x electric field
F = q x E
[tex]a = \frac{qE}{m}[/tex]
[tex]a = \frac{1.6\times 10^{-19}\times 6.325\times 10^{8}}{9.1\times 10^{-31}}[/tex]
a = 1.11 x 10^20 m/s^2
which cruising altitude is appropriate for vfr flight on a magnetic course of 135°
Answer:odd thousands plus 500 feet.
Explanation:
On a magnetic course of zero through 179 degrees, select an odd thousand foot cruising altitude plus 500 feet, such as 3,500, 5,500, up to and including 17,500. Even and odd thousands are reserved for those aircraft on an active instrument flight plan. Even thousands plus 500 feet are for aircraft flying between 180 and 359 degrees.
How long would it take for 3.5 C of charge to pass through a cross-sectional area of a wire in which a current of 5 mA passes?
Answer:
The time taken for the charge to pass through the wire = 700 s or 11.67 minutes or 0.194 hours
Explanation:
Electric Charge: This is defined as the product of electric current to time in a an electric circuit. The S.I unit of charge is Coulombs (C).
Mathematically, it is represented as
Q = it.......................... Equation 1
Where Q =quantity of charge, I = current, t = time
making t the subject of formula in equation 1,
t = Q/I ....................... Equation 2
Given: Q = 3.5 C, I = 5 mA.
Conversion: We convert from mA to A
I.e 5 mA = (5 × 10⁻³) A = 0.005 A.
Substituting these values into equation 2
t = 3.5/0.005
t = 700 seconds
or
(700/60) minutes = 11.67 minutes
or
(700/3600) hours = 0.194 hours.
Therefore the time taken for the charge to pass through the wire = 700 s or 11.67 minutes or 0.194 hours
700 second will take for 3.5 C of charge to pass through a cross-sectional area of a wire in which a current of 5 mA passes.
Partical charge and current based problem:What information do we have?
Charges of partical = 3.5 C
Current = 5 mA = (5 × 10⁻³) A
Charge = (Curernt)(time)
Q = iT
T = Q / i
Time taken = 3.5 / (5 × 10⁻³)
Time taken = 700 seconds or 11.67 minutes
Find out more information about 'Current'.
https://brainly.com/question/23323183?referrer=searchResults
Poiseuille's law remains valid as long as the fluid flow is laminar. For sufficiently high speed, however, the flow becomes turbulent, even if the fluid is moving through a smooth pipe with no restrictions. It is found experimentally that the flow is laminar as long as the Reynolds Number Re is less than about 2000: Re = 2v Normal 0 false false false IN X-NONE X-NONE MicrosoftInternetExplorer4 rhoR Normal 0 false false false IN X-NONE X-NONE MicrosoftInternetExplorer4 /η. Here Normal 0 false false false IN X-NONE X-NONE MicrosoftInternetExplorer4 v, rho, and η are, respectively, the average speed, density, and viscosity of fluid, and R is the radius of the pipe. Calculate the highest average speed that blood (rho = 1060 kg/m3, η = 4.0 x 10-3 Pa.s) could have and still remain in laminar flow when it flows through the aorta (R = 8.0 x 10-3 m)
Normal 0 false false false IN X-NONE X-NONE MicrosoftInternetExplorer4
Answer: 0.471 m/s
Explanation:
We are given the followin equation:
[tex]Re=\frac{D v \rho}{\eta}[/tex] (1)
Where:
[tex]Re[/tex] is the Reynolds Number, which is adimensional and indicates if the flow is laminar or turbulent
When [tex]Re<2100[/tex] we have a laminar flow
When [tex]Re>4000[/tex] we have a turbulent flow
When [tex]2100<Re<4000[/tex] the flow is in the transition region
[tex]D=2R[/tex] is the diameter of the pipe. If the pipe ha a radius [tex]R=8(10)^{-3} m[/tex] its diameter is [tex]D=2(8(10)^{-3} m)=0.016 m[/tex]
[tex]v[/tex] is the average speed of the fluid
[tex]\rho=1060 kg/m^{3}[/tex] is the density of the fluid
[tex]\eta=4(10)^{-3} Pa.s[/tex] is the viscosity of the fluid
Isolating [tex]v[/tex]:
[tex]v=\frac{Re \eta}{D \rho}[/tex] (2)
Solving for [tex]Re=2000[/tex]
[tex]v=\frac{(2000)(4(10)^{-3} Pa.s)}{(0.016 m)(1060 kg/m^{3})}[/tex] (3)
Finally:
[tex]v=0.471 m/s[/tex]
The moon orbits the earth at a distance of 3.85 x 10^8 m. Assume that this distance is between the centers of the earth and the moon and that the mass of the earth is 5.98 x 10^24 kg. Find the period for the moon's motion around the earth. Express the answer in days and compare it to the length of a month.
Answer:
27.5 days
0.92 month
Explanation:
[tex]r[/tex] = radius of the orbit of moon around the earth = [tex]3.85\times10^{8} m[/tex]
[tex]M[/tex] = Mass of earth = [tex]5.98\times10^{24} m[/tex]
[tex]T[/tex] = Time period of moon's motion
According to Kepler's third law, Time period is related to radius of orbit as
[tex]T^{2} = \frac{4\pi ^{2} r^{3} }{GM}[/tex]
inserting the values, we get
[tex]T^{2} = \frac{4(3.14)^{2} (3.85\times10^{8})^{3} }{(6.67\times10^{-11})(5.98\times10^{24})}\\T = 2.3754\times10^{6} sec[/tex]
we know that
1 day = 24 hours = 24 x 3600 sec = 86400 s
[tex]T = 2.3754\times10^{6} sec \frac{1 day}{86400 sec} \\T = 27.5 days[/tex]
1 month = 30 days
[tex]T = 27.5 days \frac{1 month}{30 days} \\T = 0.92 month[/tex]
Final answer:
The period for the moon's motion around the earth is approximately 0.59 days, which is much shorter than the length of a month.
Explanation:
To find the period for the moon's motion around the earth, we can use Kepler's third law. According to Kepler's third law, the square of the period of a planet's orbit is directly proportional to the cube of its average distance from the center of the orbit.
We are given that the moon orbits the earth at a distance of 3.85 x 10^8 m. We can use this information to calculate the period as follows:
Convert the given distance to meters: 3.85 x 10^8 m.Calculate the period using Kepler's third law equation:Hence, the period for the moon's motion around the earth is approximately 0.59 days. This is much shorter than the length of a month, which is about 30 days. Therefore, the moon completes multiple orbits around the earth in one month.
a bike travels at a constant speed for 4.00 m/s for 5.00 seconds. How far does it go
Answer:
20 metres
Explanation:
Speed = distance ÷ time
[tex]s = \frac{d}{t} [/tex]
If we substitute the values:
[tex]4 = \frac{d}{5} [/tex]
[tex]20 = d[/tex]
Answer: 20m
Explanation:
Speed = Distance/time taken
Speed = 4.00m/s
Time taken = 5.00s
Distance = D = ?
We insert the values in the formula
4.00m/s = D/5.00s
Multiply through by 5
D = 20m
A projector is placed on the ground 22 ft. away from a projector screen. A 5.2 ft. tall person is walking toward the screen at a rate of 3 ft./sec. How fast is the height of the person's shadow changing when the person is 13 ft. from the projector
Answer:
y = 67.6 feet, y = 114.4/ (22 - 3t)
Explanation:
For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram
Large triangle Projector up to the screen
tan θ = y / L
For the small triangle. Projector up to the person
tan θ = y₀ / (L-d)
The angle is the same, so we equate the two equations
y₀ / (L -d) = y / L
y = y₀ L / (L-d)
The distance from the screen (d), we look for it with kinematics
v = d / t
d = v t
we replace
y = y₀ L / (L - v t)
y = 5.2 22 / (22 - 3 t)
y = 114.4 (22 - 3t)⁻¹
This is the equation of the shadow height change as a function of time
For the suggested distance the shadow has a height of
y = 114.4 / (22-13)
y = 67.6 feet
A 5.0-kg cannonball is fired over level ground with a velocity of 2.00 ⨯ 102 m/s at an angle of 25° above the horizontal. Just before it hits the ground its speed is 150 m/s. Over the entire trip, find the change in the thermal energy of the cannonball and air.
Answer:
E=147898.01J
Explanation:
A 5.0-kg cannonball is fired over level ground with a velocity of 2.00 ⨯ 102 m/s at an angle of 25° above the horizontal. Just before it hits the ground its speed is 150 m/s. Over the entire trip, find the change in the thermal energy of the cannonball and air
firstly , we look for the time of flight it takes to make the projectile path
T=2Usin∅/g
take g= 9.81m/s
T=2*200sin25/(9.81)
T=17.23Secs
energy is force *distance
E=f*d
f=m*g
f=5*9.81
f=49.05N
s=distance
s=(v+u)T/2
s=(150+200)17.23/2
s=3015.25m
49.05N*3015.25m
E=147898.01J
Which statement is true regarding radioactive particles subjected to an electric field?
Answer:There are three types of radiation
Alpha, Beta and Gamma radiation
Explanation: In an electric field produced by two parallel charged plates alpha particle would be deflected toward a - plate following a parabolic path, beta rays toward a +plate following a parabolic path and gamma radiation either - or + source.
The drawing shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface (1) has an area of 1.70 m², while surface (2) has an area of 4.00 m². The electric field in the drawing is uniform and has a magnitude of 210 N/C. Find the magnitude of the electric flux through surface (1) if the angle θ made between the electric field with surface (2) is 34.0°.
Answer:
Ф1=295.96Nm^2/C
Ф2=469.73Nm^2/C
Explanation:
The drawing shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface (1) has an area of 1.70 m², while surface (2) has an area of 4.00 m². The electric field in the drawing is uniform and has a magnitude of 210 N/C. Find the magnitude of the electric flux through surface (1) if the angle θ made between the electric field with surface (2) is 34.0
we have two surfaces where we know angle of surface 1 lets call it
s1= 34.
Therefore to find s2
s2 (the angle from surface 2) we have that
s2=180-(90+s1),
so s2=180-(90+34),
thus s2=56 degrees.
Flux equation reads as Φ=ΕΑ,
where Φ is the flux,
E is the electric field and
A is the surface area.
So with respect to the angles and the figure provided,
we have Φ=EAcos(s).
So we can solve further by writing
. For Surface 1 we have
Φ1=EAcos(s1)=210 x 1.7 x cos 34,
so Φ1=295.96,
approximately to an whole number
Φ1=296 Nm^2/c.
Similarly for Φ2, we have
Φ2=EAcos(s2)
=210 x 4 x cos34=469.7,
thus Φ2=469.7Nm^2/c.
What type of wave is shown above?
A. transverse wave
B. longitudinal wave
C. surface wave
D. electromagnetic wave
Answer:
its a A. Transverse wave