Answer:
3900000 m/s
Explanation:
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
v = Speed of the galaxy relative to the earth
Observed frequency is
[tex]f'=(1+0.013)f\\\Rightarrow \dfrac{f'}{f}=1.013[/tex]
Here the Doppler relation must be used.
So, observed frequency is given by
[tex]f'=f\dfrac{v+c}{c}\\\Rightarrow \dfrac{f'}{f}=\dfrac{v+c}{c}\\\Rightarrow v=\dfrac{f'c}{f}-c\\\Rightarrow v=(1.013\times 3\times 10^8)-3\times 10^8\\\Rightarrow v=3900000\ m/s[/tex]
The speed of the galaxy relative to the earth is 3900000 m/s
A gas in a cylinder is held at a constant pressure of 1.80 * 105 Pa and is cooled and compressed from 1.70 m3 to 1.20 m3 . The internal energy of the gas decreases by 1.40 * 105 J.
(a) Find the work done by the gas.
(b) Find the absolute value |Q| of the heat flow into or out of the gas, and state the direction of the heat flow.
(c) Does it matter whether the gas is ideal? Why or why not
This question is wrong because of not correct values.The Correct question is
A gas in a cylinder is held at a constant pressure of 1.80×10⁵Pa and is cooled and compressed from 1.70m³ to 1.20m³. The internal energy of the gas decreases by 1.40×10⁵J.
(a) Find the work done by the gas.
(b) Find the absolute value of the heat flow, |Q|, into or out of the gas, and state the direction of the heat flow.
(c) Does it matter whether the gas is ideal? Why or why not?
Answer:
(a) W= -9×10⁴J
(b) |Q|=2.3×10⁵
(c) It does not matter whether gas is ideal or non-ideal
Explanation:
Given
V₁=1.70m³
V₂=1.20m³
p=1.8×10⁵ pa
ΔU= -1.40×10⁵J
For (a) work done by gas
[tex]W=p(V_{2}-V_{1} )\\W=1.8*10^{5}(1.2-1.7)\\ W=-9*10^{4}J[/tex]
For (b) Heat flow |Q|
|Q|=ΔU+W
[tex]Q=(-1.4*10^{5}) +(-9*10^{4})\\Q=-2.3*10^{5}J\\So\\|Q|=|-2.3*10^{5}|\\|Q|=2.3*10^{5}J[/tex]
For (c) part
It does not matter whether the gas is ideal or not because the first law of thermodynamics which applied in our solution could applied to any material ideal or non ideal
A 0.6-m3 rigid tank contains 0.6 kg of N2 and 0.4 kg of O2 at 300 K. Determine the partial pressure of each gas and the total pressure of the mixture. The gas constant for N2 is 0.2968 kPa·m3/kg·K and the gas constant for O2 is 0.2598 kPa·m3/kg·K.
Answer:
Pnitrogen=3.18 kPa, Poxygen=1.62 kPa , Ptotal= 4.80 kPa
Explanation:
partial pressure equation becomes Ptotal = Pnitrogen + Poxygen
Partial pressure of Nitrogen
Pnitrogen= nRT/V
n=no of moles =mass/molar mass
mass of nitrogen=0.6kg
Molar mass of nitrogen gas=28gmol^-1
n=0.6/28=0.0214moles
R=0.2968 kPa·m3/kg·K
T=300k
V=0.6m^3
Pnitrogen=(0.0214 * 0.2968 * 300)/0.6
Pnitrogen=3.18 kPa
Likewise
Poxygen=nRT/V
n=0.4/32=0.0125moles
R=0.2598 kPa·m3/kg·K
T=300k
V=0.6m^3
Poxygen=(0.0125 * 0.2598 * 300)/0.6
Poxygen=1.62 kPa
Ptotal= 3.18+1.62= 4.80 kPa
Using the ideal gas law, the partial pressure for Nitrogen ([tex]N_{2}[/tex]) and Oxygen ([tex]O_{2}[/tex]) is 148.4 kPa and 86.6 kPa respectively. The total pressure of the mixture is the sum of these two partial pressures, equaling 235 kPa.
Explanation:The pressure exerted by individual gases in a mixture is known as partial pressure. The calculation of the partial pressure of each gas, and the total pressure of the mixture involves using the ideal gas law. In this case, the ideal gas equation is P = mRT/V, where P represents the pressure, m is the mass of gas, R is the gas constant, T is the temperature and V is the volume. Thus, for Nitrogen ([tex]N_{2}[/tex]), the partial pressure is (0.6 kg × 0.2968 kPa·m3/kg·K × 300K) / 0.6 m3 = 148.4 kPa, and for Oxygen ([tex]O_{2}[/tex]), the partial pressure is (0.4 kg× 0.2598 kPa·m3/kg·K × 300K) / 0.6 m3 = 86.6 kPa. The total pressure, then, is the sum of these partial pressures, which equals 148.4 kPa + 86.6 kPa = 235 kPa.
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A titanium (c = 520 J/kg*K) satellite of mass m = 500 kg at a temperature of 10 K is in geostationary orbit above the equator. It is impacted by a meteorite, which knocks it off course but does not affect its speed. Because of the change in trajectory, the satellite will now crash-land on Earth in a small pond containing 5 × 10^5 kg of water (c = 4186 J/kg*K) at 20°C. Ignoring air resistance, and assuming that the water stays in the pond somehow, what is the final temperature of the water? Please indicate if any water boils, and justify any simplifying assumptions you make in your solution.
A tube has a length of 0.025 m and a cross-sectional area of 6.5 x 10-4 m2. The tube is filled with a solution of sucrose in water. The diffusion constant of sucrose in water is 5.0 x 10-10 m2/s. A difference in concentration of 5.2 x 10-3 kg/m3 is maintained between the ends of the tube. How much time is required for 5.7 x 10-13 kg of sucrose to be transported through the tube?
Answer:
The time required for sucrose transportation through the tube is 8.4319 sec.
Explanation:
Given:
L = 0.025 m
A = 6.5×10^-4 m^2
D = 5×10^-10 m^2/s
ΔC = 5.2 x 10^-3 kg/m^3
m = 5.7×10^-13 kg
Solution:
t = m×L / D×A×ΔC
t = (5.7×10^-13) × (0.025) / (5×10^-10)×(6.5×10^-4)×(5.2 x 10^-3)
t = 8.4319 sec.
As a comet approaches the Sun, it arcs around the Sun more rapidly. Why?
Answer:
Explanation:
According to Kepler's law a radius vector joining planet and sun swept equal area in equal interval of time thus it can be applied for comets.
when a comet is nearer to sun it has to swept more area so its velocity is more nearer to the sun ,
The basics of this formula comes from conservation of angular momentum thus comet moves faster when it approaches the sun.
Light from a laser strikes a diffraction grating that has 5 300 grooves per centimeter. The central and first-order principal maxima are separated by 0.488 m on a wall 1.64 m from the grating. Determine the wavelength of the laser light. (In this problem, assume that the light is incident normally on the gratings.)
To solve this problem it is necessary to apply the concepts related to the principle of superposition, as well as to constructive interference. From the definition we know that this can be expressed mathematically as
[tex]sin\theta_m = \frac{m\lambda}{d}[/tex]
Where
m = Any integer which represent the number of repetition of spectrum
[tex]\lambda[/tex]= Wavelength
d = Distance between slits
From triangle (Watch image below)
[tex]tan\theta_1 = \frac{y_1}{L}[/tex]
[tex]tan\theta_1 = \frac{0.488}{1.64}[/tex]
[tex]\theta_1 = 16.57\°[/tex]
Replacing the angle at the first equation for m=1 we have
[tex]\lambda = d sin \theta_1[/tex]
Each of the distances (d) would be defined by
d = \frac{1}{5300} = 0.0001886
[tex]\lambda = (\frac{1}{5300}) sin(16.57)[/tex]
[tex]\lambda = 538nm[/tex]
Therefore the wavelength of the laser light is 538nm
A 946.4 kg ( 2083 lb ) car is moving at 14.8 m / s ( 33.0 mph ) . Calculate the magnitude of its momentum.
Answer:
14009. 72 kgms^-1
Explanation:
Momentum is the product of an objects mass and velocity
The calculated value of the magnitude of the car's momentum be 14006.72 kg.m/s or 68739 lb.mph.
What is momentum?Momentum is a property of a body moving in linear motion, It is product of mass and velocity. As velocity is a vector quantity, momentum is also a vector quantity with magnitude and direction. SI unit of momentum be kg.m/s.
Given parameters:
Mass of the car: m = 946.4 kg. = 2083 lb
Speed of the car: v = 14.8 m/s. = 33.0 mph
So, momentum of the car: p = mv = 946.4 × 14.8 kg.m/s = 14006.72 kg.m/s.
or p = mv = 2083 lb×33.0 mph = 68739 lb.mph.
Hence, the magnitude of its momentum be 14006.72 kg.m/s or 68739 lb.mph.
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A 5 kg object near Earth's surface is released from rest such that it falls a distance of 10 m. After the object falls 10 m, it has a speed of 12 m/s. Which of the following correctly identifies whether the object-Earth system is open or closed and describes the net external force?
A. The system is closed, and the net external force is zero.
B. The system is open, and the net external force is zero.
C. The system is closed, and the net external force is nonzero.
D. The system is open, and the net external force is nonzero.
Answer:D
Explanation:
Given
mass of object [tex]m=5 kg[/tex]
Distance traveled [tex]h=10 m[/tex]
velocity acquired [tex]v=12 m/s[/tex]
conserving Energy at the moment when object start falling and when it gains 12 m/s velocity
Initial Energy[tex]=mgh=5\times 9.8\times 10=490 J[/tex]
Final Energy[tex]=\frac{1}{2}mv^2+W_{f}[/tex]
[tex]=\frac{1}{2}\cdot 5\cdot 12^2+W_{f}[/tex]
where [tex]W_{f}[/tex] is friction work if any
[tex]490=360+W_{f}[/tex]
[tex]W_{f}=130 J[/tex]
Since Friction is Present therefore it is a case of Open system and net external Force is zero
An open system is a system where exchange of energy and mass is allowed and Friction is acting on the object shows that system is Open .
The second law of thermodynamics states that spontaneous processestend to be accompanied by entropy increase. Consider, however, thefollowing spontaneous processes:
the growth of plants from simple seeds to well-organizedsystems
the growth of a fertilized egg from a single cell to a complexadult organism
the formation of snowflakes from molecules of liquid water withrandom motion to a highly ordered crystal
the growth of organized knowledge over time
In all these cases, systems evolve to a state of less disorder andlower entropy, apparently violating the second law ofthermodynamics. Could we, then, consider them as processesoccurring in systems that are not isolated?
True or False?
Answer:
True
Explanation:
According to the definition of a closed system, It is true because it's not precisely a closed system. A closed system is a physical system that doesn't let certain types of transfers (such as transfer of mass in or out of the system), though the transfer of energy is allowed.
A miniature spring-loaded, radio-controlled gun is mounted on an air puck. The gun's bullet has a mass of 7.00 g, and the gun and puck have a combined mass of 150 g. With the system initially at rest, the radio controlled trigger releases the bullet causing the puck and empty gun to move with a speed of 0.530 m/s. What is the bullet's speed?
Answer:
Explanation:
Given
mass of bullet [tex]m=7 gm[/tex]
mass of gun and Puck [tex]M=150 gm[/tex]
speed of gun and Puck is [tex]v=0.53 m/s[/tex]
Let speed of bullet be u
conserving Momentum
initial momentum=Final Momentum
[tex]0=Mv+mu[/tex]
[tex]u=-\frac{M}{m}v[/tex]
[tex]u=-\frac{150}{7}\times 0.53=-11.35 m/s[/tex]
negative sign indicates that bullet is moving in opposite direction
Using the law of conservation of momentum, it's possible to calculate that the bullet moves in the opposite direction of the gun and puck at a speed of approximately 113.6 m/s.
Explanation:This question represents a practical example of the conservation of momentum. According to the law of conservation of momentum, the total momentum of an isolated system remains constant if no external forces act on it. As the system (gun and bullet) is initially at rest, the total momentum is zero, and it remains zero even after the bullet is fired.
Let's denote the bullet's velocity as v. The momentum of the bullet after firing is its mass times its velocity, or 0.007kg * v. The momentum of the gun and puck is their combined mass times their velocity, or 0.150kg * 0.530m/s. As the total momentum should remain zero, these two quantities must be equal and opposite. Thus, 0.007kg * v = - 0.150kg * 0.530m/s. Solving for v, we find that the bullet's speed is approximately -113.6 m/s. The negative sign indicates that the bullet moves in the opposite direction of the gun and puck.
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An engineer has an odd-shaped 13.5 kg object and needs to find its rotational inertia about an axis through its center of mass. The object is supported on a wire stretched along the desired axis. The wire has a torsion constant κ = 0.618 N·m. If this torsion pendulum oscillates through 28 cycles in 58.1 s, what is the rotational inertia of the object?
Answer:
I = 0.0674 kg.m²
Explanation:
given,
mass = 13.5 Kg
torsion constant = k = 0.618 N.m
number of cycle = 28
time = 58.1 s
Time of one cycle
[tex]T = \dfrac{58.1}{28}[/tex]
[tex]T =2.075\ s[/tex]
we know,
[tex]T = 2\pi\sqrt{\dfrac{I}{k}}[/tex]
[tex]I = k (\dfrac{T}{2\pi})^2[/tex]
[tex]I =0.618\times \dfrac{T^2}{4\pi^2}[/tex]
[tex]I =0.618\times \dfrac{2.075^2}{4\pi^2}[/tex]
I = 0.0674 kg.m²
the rotational inertia of the object is equal to I = 0.0674 kg.m²
An average nerve axon is about 4x10-6m in radius, and the axoplasm that composes the interior of the axon has a resistivity of about 2.3 Ω ⋅m.
What is the resistance of just 2-cm length of this axon?
Provide your answer in mega-ohms (1 mega-ohm = 106 ohms or "millions of ohms").
(Note, this value is so large -- it corresponds to the resistance of tens of thousands of miles of the thinnest copper wire normally manufactured(!) -- that it explains why a nerve pulse traveling down an axon CANNOT simply be a current traveling along the axon. The voltage required to achieve a perceptible current in the axon would have to be gigantic! We will investigate how voltage pulses -- not current -- travel down axons in a future lab.)
For the calculation of resistance there are generally two paths. The first is through Ohm's law and the second is through the relationship
[tex]R = \frac{pL}{A}[/tex]
Where
p = Specific resistance of material
L = Length
A = Area
The area of nerve axon is given as
[tex]A = \pi r^2[/tex]
[tex]A = \pi (5*10^{-6})^2[/tex]
[tex]A = 7.854*10^{-11}m^2[/tex]
The rest of values are given as
[tex]p= 2 \Omega\cdot m[/tex]
[tex]L = 2cm = 0.02m[/tex]
Therefore the resistance is
[tex]R = \frac{pL}{A}[/tex]
[tex]R = \frac{2*0.02}{7.854*10^{-11}}[/tex]
[tex]R = 509.3*10^6\Omega[/tex]
[tex]R = 509.3M\Omega[/tex]
The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers.When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.At what speed v should an archerfish spit the water to shoot down an insect floating on the water surface located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle 60∘ above the water surface.
Answer:
Explanation:
Here is the full question and answer,
The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers.
When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.
Part A: At what speed v should an archerfish spit the water to shoot down a floating insect located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle 60 degrees above the water surface.
Part B: Now assume that the insect, instead of floating on the surface, is resting on a leaf above the water surface at a horizontal distance 0.600 m away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle 60 degrees above the surface and with the same initial speed v as before. At what height h above the surface was the insect?
Answer
A.) The path of a projectile is horizontal and symmetrical ground. The time is taken to reach maximum height, the total time that the particle is in flight will be double that amount.
Calculate the speed of the archer fish.
The time of the flight of spitted water is,
[tex]t = \frac{{2v\sin \theta }}{g}[/tex]
Substitute [tex]9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}[/tex] for g and [tex]60^\circ[/tex] for [tex]\theta[/tex] in above equation.
[tex]t = \frac{{2v\sin 60^\circ }}{{9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}}}\\\\ = \left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\[/tex]
Spitted water will travel [tex]0.80{\rm{ m}}[/tex] horizontally.
Displacement of water in this time period is
[tex]x = vt\cos \theta[/tex]
Substitute [tex]\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}[/tex] for [tex]t\rm 60^\circ[tex] for [tex]\theta[/tex] and [tex]0.80{\rm{ m}}[/tex] for x in above equation.
[tex]\\0.80{\rm{ m}} = v\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\left( {\cos 60^\circ } \right)\\\\0.80{\rm{ m}} = {v^2}\left( {0.1767{\rm{ }}} \right)\frac{1}{2}{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\\\v = \sqrt {\frac{{2\left( {0.80{\rm{ m}}} \right)}}{{0.1767\;{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}}}} \\\\ = 3.01{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\[/tex]
B.) There are two component of velocity vertical and horizontal. Calculate vertical velocity and horizontal velocity when the angle is given than calculate the time of flight when the horizontal distance is given. Value of the horizontal distance, angle and velocity are given. Use the kinematic equation to solve the height of insect above the surface.
Calculate the height of insect above the surface.
Vertical component of the velocity is,
[tex]{v_v} = v\sin \theta[/tex]
Substitute [tex]3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for v and [tex]60^\circ[/tex] for [tex]\theta[/tex] in above equation.
[tex]\\{v_v} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\sin 60^\circ \\\\ = 2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\[/tex]
Horizontal component of the velocity is,
[tex]{v_h} = v\cos \theta[/tex]
Substitute [tex]3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for v and [tex]60^\circ[/tex] for [tex]\theta[/tex] in above equation.
[tex]\\{v_h} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\cos 60^\circ \\\\ = 1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\[/tex]
When horizontal [tex]({0.60\;{\rm{m}}}[/tex] distance away from the fish.
The time of flight for distance (d) is ,
[tex]t = \frac{d}{{{v_h}}}[/tex]
Substitute [tex]0.60\;{\rm{m}}[/tex] for d and [tex]1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for [tex]{v_h}[/tex] in equation [tex]t = \frac{d}{{{v_h}}}[/tex]
[tex]\\t = \frac{{0.60\;{\rm{m}}}}{{1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}}}\\\\ = 0.3987{\rm{ s}}\\[/tex]
Distance of the insect above the surface is,
[tex]s = {v_v}t + \frac{1}{2}g{t^2}[/tex]
Substitute [tex]2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for [tex]{v_v}[/tex] and [tex]0.3987{\rm{ s}}[/tex] for t and [tex]- 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}[/tex] for g in above equation.
[tex]\\s = \left( {2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\left( {0.3987{\rm{ s}}} \right) + \frac{1}{2}\left( { - 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right){\left( {0.3987{\rm{ s}}} \right)^2}\\\\ = 0.260{\rm{ m}}\\[/tex]
A peg is located a distance h directly below the point of attachment of the cord. If h = 0.760 L, what will be the speed of the ball when it reaches the top of its circular path about the peg?
Answer:
Explanation:
Given
Pivot is at h=0.76 L
Where L is the length of String
Conserving Energy at A and B
[tex]mgL=\frac{1}{2}mu^2[/tex]
where u=velocity at bottom
[tex]u=\sqrt{2gL}[/tex]
After coming at bottom the ball completes the circle with radius r=L-0.76 L
Suppose v is the velocity at the top
Conserving Energy at B and C
[tex]\frac{1}{2}mu^2=mg(2r)+\frac{1}{2}mv^2[/tex]
Eliminating m
[tex]u^2=4r+v^2[/tex]
[tex]v^2=u^2-4\cdot gr[/tex]
[tex]v^2=2gL-4g(L-0.76L)[/tex]
[tex]v^2=1.04gL[/tex]
[tex]v=\sqrt{1.04gL}[/tex]
Which is true about the self-induced emf of an inductor?
It is a fixed value, depending on only the geometry of the device.
It depends on the amount of current through the inductor.
It depends on the rate of dissipation.
It depends on the rate at which the current through it is changing.
Answer:
It depends on the rate at which the current through it is changing.
Explanation:
As per the Faraday's law, the induced emf is given by :
[tex]\epsilon=-L\dfrac{di}{dt}[/tex]
Where
L is the inductance of the inductor
[tex]\dfrac{di}{dt}[/tex] is the rate of change of current
So, the self-induced emf of an inductor depends on the rate at which the current through it is changing. Hence, the correct option is (d).
You want to apply 18 N ⋅ m of torque to tighten a nut, using a wrench with a length of either 0.400 m, 0.600 m, or 0.900 m. For each wrench, determine the perpendicular force that you would have to apply to the end of the wrench in order to produce the desired torque.
Answer:
45 N, 30N ,20N
Explanation
torque = perpendicular force× distance
perpendicular force = torque/ distance
1) perpendicular force = 18/0.400 = 45 N
2) perpendicular force = 18/0.600 = 30N
3) perpendicular force = 18/0.900 = 20N
We have that for the Question " the perpendicular force that you would have to apply to the end of the wrench in order to produce the desired torque" it can be said that
F=45NF=30NF=20N
From the question we are told
You want to apply 18 N ⋅ m of torque to tighten a nut, using a wrench with a length of either 0.400 m, 0.600 m, or 0.900 m. For each wrench, determine the perpendicular force that you would have to apply to the end of the wrench in order to produce the desired torque.
Generally the equation for the Force is mathematically given as
t=F*d
a)
[tex]t=F*d\\\\F=\frac{t}{d}\\\\F=\frac{18}{0.4}\\\\[/tex]
F=45N
b)
[tex]t=F*d\\\\F=\frac{t}{d}\\\\\F=\frac{18}{0.6}\\\\[/tex]
F=30N
c)
[tex]t=F*d\\\\F=\frac{t}{d}\\\\F=\frac{18}{0.6}\\\\[/tex]
F=20N
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A Machine part is undergoing SHM with a frequency of 5.00 Hz and amplitude 1.80 CM, how long does it take the part to go from x=0 to x=-1.80 cm?
It takes 0.05 seconds for a part undergoing simple harmonic motion with a frequency of 5.00 Hz to move from x=0 to x=-1.80 cm.
Explanation:The student is asking how long it takes for a machine part undergoing simple harmonic motion (SHM) with a frequency of 5.00 Hz and amplitude of 1.80 cm to move from its equilibrium position (x=0) to its maximum negative displacement (x=-1.80 cm).
In SHM, the period (T) is the time required for one complete cycle of the motion. The period can be calculated using the formula T = 1/f, where f is the frequency. For a frequency of 5.00 Hz, the period is T = 1/5.00 Hz = 0.20 s. In half of a period, the part moves from one extreme to the other, passing through the equilibrium position at the quarter-period. Thus, to go from x=0 to x=-1.80 cm, it takes a quarter of this period, or (1/4)T = 0.05 s.
Therefore, it takes 0.05 seconds for the part to travel from x=0 to x=-1.80 cm while undergoing SHM.
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In a football game, a 90 kg receiver leaps straight up in the air to catch the 0.42 kg ball the quarterback threw to him at a vigorous 21 m/s, catching the ball at the highest point in his jump. Right after catching the ball, how fast is the receiver moving?
To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as
[tex]m_1v_1+m_2v_2 = (m_1+m_2)v_f[/tex]
Where,
[tex]m_{1,2}[/tex]= Mass of each object
[tex]v_{1,2}[/tex] = Initial velocity of each object
[tex]v_f[/tex]= Final Velocity
Since the receiver's body is static for the initial velocity we have that the equation would become
[tex]m_2v_2 = (m_1+m_2)v_f[/tex]
[tex](0.42)(21) = (90+0.42)v_f[/tex]
[tex]v_f = 0.0975m/s[/tex]
Therefore the velocity right after catching the ball is 0.0975m/s
The receiver's final velocity is approximately 0.098 m/s.
Explanation:First, we need to find the initial momentum of the ball by multiplying its mass (0.42 kg) by its velocity (21 m/s). The initial momentum of the ball is therefore 8.82 kg·m/s. Next, we need to find the final momentum of the receiver by multiplying his mass (90 kg) by his final velocity. Since he catches the ball at the highest point in his jump, his final velocity is 0 m/s. So the final momentum of the receiver is 0 kg·m/s. According to the law of conservation of momentum, the initial momentum of the ball must be equal to the final momentum of the receiver. Therefore, the final velocity of the receiver is 8.82 kg·m/s divided by his mass (90 kg), which is approximately 0.098 m/s.
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Choose the scenario under which each of the given Doppler shift effects will be seen.a. The source and observer are approaching one anotherb. The source and observer are moving away from one anotherc. The source and observer are stationary relative to one another
Answer:
scenario A and B
Explanation:
The Doppler effect is the change in frequency of a wave by the relative movement of the source and the observer. It is described by the expression
f ’= (v + v₀) / (v - [tex]v_{s}[/tex]) f₀
Where f₀ is the emitted frequency, v the speed of the wave, v₀ and [tex]v_{s}[/tex] the speed of the observer and the source, respectively, the signs are for when they are approaching and in the case of being away the signs are changed.
Consequently, from the above for the Doppler effect to exist there must be a relative movement of the source and the observer.
Let's examine the scenarios
A) True. You agree with the equation shown
B) True. Only the signs should be changed and it is described by the equation shown
C) False. If there is no relative movement there is no Doppler effect
You have 10 ohm and a 100 ohm resistor in parallel. You place this equivalent resistance in series with an LED, which is rated to have a voltage drop of 1.83V at 20mA. What voltage should you connect to this circuit to provide 20mA of current for the LED? What would the current be, if you connected to a 3V battery? Draw all of these diagrams before attempting the calculations.
Answer:
Approximately [tex]\rm 2.0\; V[/tex].
Approximately [tex]\rm 30 \; mA[/tex]. (assumption: the LED here is an Ohmic resistor.)
Explanation:
The two resistors here [tex]R_1= 10\; \Omega[/tex] and [tex]R_2= 100\; \Omega[/tex] are connected in parallel. Their effective resistance would be equal to
[tex]\displaystyle \frac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2}} = \frac{1}{\dfrac{1}{10} + \dfrac{1}{100}} = \frac{10}{11} \; \Omega[/tex].
The current in a serial circuit is supposed to be the same everywhere. In this case, the current through the LED should be [tex]20\; \rm mA = 0.020\; \rm A[/tex]. That should also be the current through the effective [tex]\displaystyle \rm \frac{10}{11} \; \Omega[/tex] resistor. Make sure all values are in standard units. The voltage drop across that resistor would be
[tex]V = I \cdot R = 0.020 \times \dfrac{10}{11} \approx 0.182\; \rm V[/tex].
The voltage drop across the entire circuit would equal to
the voltage drop across the resistors, plusthe voltage drop across the LED.In this case, that value would be equal to [tex]1.83 + 0.182 \approx 2.0\; \rm V[/tex]. That's the voltage that needs to be supplied to the circuit to achieve a current of [tex]20\; \rm mA[/tex] through the LED.
Assuming that the LED is an Ohmic resistor. In other words, assume that its resistance is the same for all currents. Calculate its resistance:
[tex]\displaystyle R(\text{LED}) = \frac{V(\text{LED})}{I(\text{LED})}= \frac{1.83}{0.020} \approx 91.5\; \Omega[/tex].
The resistance of a serial circuit is equal to the resistance of its parts. In this case,
[tex]\displaystyle R = R(\text{LED}) + R(\text{Resistors}) = 91.5 + \frac{10}{11} \approx 100\; \Omega[/tex].
Again, the current in a serial circuit is the same in all appliances.
[tex]\displaystyle I = \frac{V}{R} = \frac{3}{100} \approx 0.030\; \rm A = 30\; mA[/tex].
The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of friction between each block and the surface is 0.4 . The two blocks collide, stick together, and move to the right. Remember that the spring is not attached to the 8 kg block. Find the speed of the 8 kg block just before it collides with the 2 kg block. Answer in units of m/s.
Answer:
3.258 m/s
Explanation:
k = Spring constant = 263 N/m (Assumed, as it is not given)
x = Displacement of spring = 0.7 m (Assumed, as it is not given)
[tex]\mu[/tex] = Coefficient of friction = 0.4
Energy stored in spring is given by
[tex]U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J[/tex]
As the energy in the system is conserved we have
[tex]\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s[/tex]
The speed of the 8 kg block just before collision is 3.258 m/s
The speed of the 8 kg block before the collision is found by considering the work done by friction and the initial potential energy in the spring and using the conservation of energy principle to solve for the speed.
Explanation:To find the speed of the 8 kg block just before it collides with the 2 kg block, we must calculate the motion of the block under the influence of friction before the collision occurs.
First, we determine the force of friction using the coefficient of friction (μ) and the normal force (N), which in this case equals the weight of the block (m × g) since the surface is horizontal. So, the force of friction is given by Fₙ = μ × m × g = 0.4 × 8 kg × 9.8 m/s² = 31.36 N.
Since there is friction, work done by friction must be considered to find the velocity. The work done by friction (Wₙ) is the force of friction (Fₙ) multiplied by the distance the block moves before colliding with the second block (d), which is not given in this problem. But if we had the distance, we could use the work-energy principle, where the net work done on an object is equal to the change in its kinetic energy (KE).
Assuming no other forces act on the block, then the work done by friction is equal to the initial potential energy stored in the spring. Using Hooke's Law, the potential energy (PE) in the spring is given by PE = ½ kx², where k is the spring constant and x is the compression distance. This energy gets converted entirely into the kinetic energy of the 8 kg block minus the losses due to friction.
Using the conservation of energy:
PE - Wₙ = ½ m v²
½ kx² - Fₙ × d = ½ m v²
We solve this equation for v, the speed of the block, if we know the distance (d).
A special system is set up in a lab that lets its user select any wavelength between 400nm and 700 nm with constant intensity. This light is directed at a thin glass film (n =1.53) with a thickness of 350 nm and that is surrounded by air. As one scans through these possible wavelengths, which wavelength of light reflected from the glass film will appear to be the brightest, if any?
a) 428 nm
b) 535 nm
c) 657 nm
d) 700 nm
e) Since the intensity of the light is constant,
all wavelengths of light reflected from the glass will appear to be the same.
Answer:
Explanation:
The case relates to interference in thin films , in which we study interference of light waves reflected by upper and lower surface of a medium or glass.
For constructive interference , the condition is
2μt = ( 2n+1)λ/2
μ is refractive index of glass , t is thickness , λ is wavelength of light.
putting the given values
2 x 1.53 x 350 x 10⁻⁹ = ( 2n+1) λ/2
λ = 2142nm / ( 2n+1)
For n = 2
λ = 428 nm
This wave length will have constructive interference making this light brightest of all .
For n = 1
λ = 714 nm
So second largest brightness will belong to 700 nm wavelength.
You set out to reproduce Thomson’s e/m experiment with an accelerating potential of 150 V and adeflecting electric field of magnitude 6.0 ×106 N/C(a) At what fraction of the speed of light do the electrons move?(b) What magnetic-field magnitude will yield zero beam deflection?(c) With this magnetic field, how will the electron beam behave if you increase the acceleratingpotential above 150 V?
Answer:
a. v' = 24.22 x 10 ⁻³
b. β = 0.825 T
Explanation:
V = 150 V , E = 6.0 x 10 ⁶ N / C
a.
¹/₂ * m * v² = e * V
v = √ 2 * e / m * V
v = √ 2 * 1.76 x 10 ¹¹ * 150 v = √ 5.28 x 10 ¹³
v = 7.26 x 10 ⁶ m /s
v' / c = 7.26 x 10 ⁶ m /s / 3.0 x 10 ⁸ = 24.22 x 10 ⁻³
b.
β = E / V
β = 6.0 x 10 ⁶ / 7.266 x 10 ⁶
β = 0.825 T
c.
When the increasing the accelerating potential speed it doesn't change the up wired the electric force and the magnetic force the electron be beat down more .
A parallel plate capacitor is being charged by a constant current i. During the charging, the electric field within the plates is increasing with time.
Which one of the following statements concerning the magnetic field between the plates is true?
A) The magnetic field within a parallel plate capacitor is always equal to zero teslas.
B) The induced magnetic field is directed antiparallel to the increasing electric field.
C) The induced magnetic field strength has its largest value at the center of the plates and decreases linearly toward the edges of the plates.
D) At a given moment, the induced magnetic field strength has the same magnitude everywhere within the plates of the capacitor, except near the edges.
E) The induced magnetic field strength is zero teslas near the center of the plates and increases as r increases toward the edges of the plates.
Answer:
E
Explanation:
The parallel plate capacitor is being charged by a steady current i. If the constant current is flowing along a straight wire, what happens inside the capacitor, between the plates is that the induced magnetic field strength is zero Tesla near the center of the plates and increases as r increases toward the edges of the plates.
From Ampère–Maxwell law, the magnetic field between the capacitor plates assuming that the capacitor is being charged at a constant rate by a steady current is
B = μ₀r /2A [tex]\frac{dQ}{dt} e_{o}[/tex]
where;
μ₀ is permeability
Q(t) is the instantaneous charge on the positive plate,
A is the cross-sectional area of a plate and
e₀ is a unit vector
A 2.0 kg block is held at rest against a spring with a force constant k = 264 N/m. The spring is compressed a certain distance and then the block is released. When the block is released, it slides across a surface that has no friction except for a 10.0 cm section that has a coefficient of friction μk = 0.54.
Find the distance in centimeters the spring was compressed such that the block's speed after crossing the rough area is 2.7 m/s.
Answer:
21.73 cm
Explanation:
We have given parameters:
Mass of block, m = 2.0 kg
Force constant of spring, k = 264 N/m
Length of rough area, L = 10 cm = 0.1 m
Co-efficient of kinetic friction , [tex]\mu_{k}[/tex] = 0.54
Block's speed after crossing rough area, v = 2.7 m/s
Block's initial speed ( when it was released from compressed spring), [tex]v_{0}[/tex] = 0 m/s
We need to find the distance that the spring was initially compressed, x = ?
Hence, we well apply Work-Energy principle which indicates that,
Work done by the friction = Change in the total energy of block
[tex]- \mu_{k} * mg * L = (\frac{1}{2} * mv^{2} - \frac{1}{2} * mv_{0} ^{2} ) + (0 - \frac{1}{2} * kx^{2})[/tex]
-0.54 * 2 * 9.8 * 0.1 = (1/2 * 2 * [tex]2.7^{2}[/tex] - 1/2 * 2 * 0) + (0 - 1/2 * 264 * [tex]x^{2}[/tex])
x = 0.2173 m = 21.73 cm
A silver bar of length 30 cm and cross-sectional area 1.0 cm2 is used to transfer heat from a 100°C reservoir to a 0°C block of ice.
How much ice is melted per second? (For silver, k = 427 J/s⋅m⋅°C. For ice, Lf = 334 000 J/kg.)
a. 4.2 g/s
b. 2.1 g/s
c. 0.80 g/s
d. 0.043 g/s
Answer:
d. 0.043 g/s
Explanation:
Formula for rate of conduction of heat through a bar per unit time is as
follows
Q = k A ( t₁ - t₂ ) / L
A is cross sectional area and L is length of rod ,( t₁ - t₂ ) is temperature
difference . Q is heat conducted per unit time
Putting the values in the equation
Q = (427 x 1 x 10⁻⁴ x 100 )/ 30 x 10⁻²
= 14.23 J/s
mass of ice melted per second
= Q / Latent heat of ice
= 14.23 / 334000
= 0.043 g/s
Final answer:
To find the mass of ice melted per second, the rate of heat transfer is calculated using the thermal conductivity of silver and the latent heat of fusion of ice. The calculated value of 0.426 g/s does not match the provided options, suggesting a possible typo in the options or the need for clarification.
Explanation:
The question is about calculating the amount of ice that is melted per second when a silver bar conducts heat from a hot reservoir to a block of ice. To answer this question, we will use the given thermal conductivity of silver (k = 427 J/s·m·°C), the cross-sectional area of the silver bar (1.0 cm2), and the latent heat of fusion for ice (Lf = 334,000 J/kg).
First, we need to calculate the rate of heat transfer (Q) from the 100°C reservoir through the silver bar to the 0°C ice using the formula:
Q = k · A · (ΔT/L),
where A is the cross-sectional area, ΔT is the temperature difference, and L is the length of the bar. Converting A to meters squared (A = 1.0 x 10-4 m2) and L to meters (L = 0.30 m), and plugging in the values:
Q = 427 J/s·m·°C · 1.0 x 10-4 m2 · 100°C / 0.30 m = 142.333 J/s.
Now, using the formula Q = mLf to find the mass of ice melted per second (m), where Q is the rate of heat transfer and Lf is the latent heat of fusion:
m = Q / Lf = 142.333 J/s / 334,000 J/kg = 0.000426 kg/s.
Converting this mass to grams (since 1 kg = 1000 g), we get:
m = 0.000426 kg/s · 1000 g/kg = 0.426 g/s.
This value is not exactly matching any of the options provided (a-d), but with rounding and considering significant figures, the closest answer would be 0.43 g/s, which is not listed amongst the options provided.
You look towards a traffic light and see a yellow light. If you were to drive towards it at near the speed of light, what color would it appear?
A. Red
B. Green
Answer:
B. Green
Explanation:
The change in wavelength that is caused when one object is moving towards another object from the perspective of the viewer is called the Doppler effect.
When objects move close to one another then wavelength reduces which is called blue shift while the opposite case causes the wavelength to increase which is called red shift.
Here, the color of the traffic light is yellow and you are getting closer to it so the wavelength should blue shift and green should appear.
Bandar Industries Berhad of Malaysia manufactures sporting equipment. One of the company’s products, a football helmet for the North American market, requires a special plastic. During the quarter ending June 30, the company manufactured 3,200 helmets, using 2,368 kilograms of plastic. The plastic cost the company $15,629. According to the standard cost card, each helmet should require 0.68 kilograms of plastic, at a cost of $7.00 per kilogram. Required: 1. What is the standard quantity of kilograms of plastic (SQ) that is allowed to make 3,200 helmets? 2. What is the standard materials cost allowed (SQ × SP) to make 3,200 helmets? 3. What is the materials spending variance? 4. What is the materials price variance and the materials quantity variance? (For requirements 3 and 4, indicate the effect of each variance by selecting "F" for favorable, "U" for unfavorable, and "None" for no effect (i.e., zero variance). Input all amounts as positive values. Do not round intermediate calculations.)
Answer:
1. 2176 kilograms of plastic 2. $15,232 3. $397 (U) 4. $947 (F) $1,344 (U)
Explanation:
Generally, cost variance analysis can be used to estimate the difference between the actual cost and the expected costs. However, if the actual cost is more than the expected cost, then the variance is said to be unfavorable and vice versa.
1. The standard quantity of kilograms of plastic (SQ) that is allowed to make 3,200 helmets?
We know that each helmet requires 0.68 kilograms of plastic. Thus, to make 3200 helmets, we will need 0.68*3200 = 2176 kilograms of plastic
2. The standard materials cost allowed (SQ × SP) to make 3,200 helmets.
We also know that each helmet costs $7.00 per kilogram. Therefore, to make 3200 helmets, the standard materials cost = $7.00*2176 = $15,232
3. The material spending variance = difference between the actual cost and the standard cost = $15,629 - $15,232 = $397 U
4. The materials price variance and the materials quantity variance?
The materials price variance is the actual cost- (standard cost per kilogram x actual number of plastic used) . Therefore:
Materials price variance = $15,629 - ($7 x 2,368 kg)
Materials price variance = $15,629 - $16,576 = ($947) F
Since the budgeted cost is relatively higher than the actual cost, the materials price variance is favorable (F) by $947.
The materials quantity variance = (Actual number of plastic used x Standard cost per kilogram) - Standard cost
Materials quantity variance = ($7 x 2,368 kg) - $15,232
Materials quantity variance = $16,576 - $15,232 = $1,344 U
Since the budgeted cost is relatively higher than the standard cost, the materials quantity variance is unfavorable (U) by $1,344.
To find the standard quantity of plastic and standard materials cost allowed, multiply the standard quantity per helmet by the number of helmets produced and multiply the standard quantity by the standard price per kilogram, respectively. The materials spending variance is the difference between the actual cost and the standard cost that should have been incurred. The materials price variance is the difference between the actual quantity of plastic used multiplied by the standard price per kilogram and the actual cost, while the materials quantity variance is the difference between the standard quantity of plastic allowed multiplied by the standard price per kilogram and the actual cost.
Explanation:To calculate the standard quantity of plastic (SQ) allowed to make 3,200 helmets, multiply the standard quantity per helmet by the number of helmets produced. In this case, SQ = 0.68 kg/helmet * 3,200 helmets = 2,176 kg. To calculate the standard materials cost allowed (SQ × SP), multiply the standard quantity by the standard price per kilogram. In this case, the materials cost allowed is $15,232 (2,176 kg * $7.00/kg).
To calculate the materials spending variance, subtract the actual cost from the standard cost that should have been incurred. In this case, the materials spending variance is $397 (Standard Cost - Actual Cost = $15,232 - $15,629). The materials price variance is calculated by subtracting the actual quantity of plastic used multiplied by the standard price per kilogram from the actual cost.
The materials quantity variance is calculated by subtracting the standard quantity of plastic allowed multiplied by the standard price per kilogram from the actual cost. In this case, the materials price variance is $7,774 (Actual Cost - Actual Quantity * Standard Price = $15,629 - 2,368 kg * $7.00/kg) and the materials quantity variance is -$2,145 (Actual Quantity * Standard Price - Standard Cost = 2,368 kg * $7.00/kg - $15,232).
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A skater has rotational inertia 4.2 kg.m2 with his fists held to his chest and 5.7 kg.m2 with his arms outstretched. The skater is spinning at 3.0 rev/s while holding a 2.5-kg weight in each outstretched hand; the weights are 76 cm from his rotation axis. If he pulls his hands in to his chest, so they’re essentially on his ro- tation axis, how fast will he be spinning?
Answer:
6.13428 rev/s
Explanation:
[tex]I_f[/tex] = Final moment of inertia = 4.2 kgm²
I = Moment of inertia with fists close to chest = 5.7 kgm²
[tex]\omega_i[/tex] = Initial angular speed = 3 rev/s
[tex]\omega_f[/tex] = Final angular speed
r = Radius = 76 cm
m = Mass = 2.5 kg
Moment of inertia of the skater is given by
[tex]I_i=I+2mr^2[/tex]
In this system the angular momentum is conserved
[tex]L_f=L_i\\\Rightarrow I_f\omega_f=I_i\omega_i\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{(5.7+2\times 2.5\times 0.76^2)3}{4.2}\\\Rightarrow \omega_f=6.13428\ rev/s[/tex]
The rotational speed will be 6.13428 rev/s
Final answer:
To solve this problem, we need to use the principle of conservation of angular momentum. The skater will be spinning at approximately 8.0 rev/s when his hands are brought to his chest.
Explanation:
To solve this problem, we need to use the principle of conservation of angular momentum. According to this principle, the initial angular momentum of the skater with his arms extended is equal to the final angular momentum when his hands are brought to his chest.
The initial angular momentum is given by the equation:
Li = Iiωi
where Li is the initial angular momentum, Ii is the initial moment of inertia, and ωi is the initial angular velocity. The final angular momentum is given by:
Lf = Ifωf
where Lf is the final angular momentum, If is the final moment of inertia, and ωf is the final angular velocity.
Since the skater is pulling his hands in to his chest, his moment of inertia decreases, and his angular velocity increases. We can set up the following equation:
Iiωi = Ifωf
Substituting the given values, we have:
(5.7 kg.m2)(2π(3.0 rev/s)) = (4.2 kg.m2)ωf
Solving for ωf, we find that the skater will be spinning at approximately 8.0 rev/s when his hands are brought to his chest.
An ideal gas goes through the following two-step process. 1) The container holding the gas has a fixed volume of 0.200 m3 while the pressure of the gas increases from 3.00×105 Pa to 4.00×105 Pa . 2) The container holding the gas is then compressed to a volume of 0.120 m3 while maintaining a constant pressure of 4.00×105 Pa .A) What is the total work done by the gas for this two-step process?
Answer:
[tex]W=-3.2\times 10^4\ J[/tex]
Explanation:
Given:
Process 1:
Volume of ideal gas is constant, [tex]V_1_i=0.2\ m^3[/tex]Initial pressure, [tex]P_1_i=3\times 10^5\ Pa[/tex]Final pressure, [tex]P_1_f=4\times 10^5\ Pa[/tex]Process 2:
Pressure of ideal gas is constant, [tex]P_2_f=4\times 10^5\ Pa[/tex]Final volume, [tex]V_2_f=0.12\ m^3[/tex]We know that the work done by an ideal gas is given as:
[tex]W=P\times (V_f-V_i)[/tex]
Now for process 1:
[tex]W_1=0\ J[/tex]
∵there is no change in volume in this process.
For process 2:
[tex]W_2=4\times 10^5\time (0.12-0.2)\ J[/tex]
[tex]W_2=-3.2\times 10^4\ J[/tex]
∵Negative , sign indicates that the work is being done on the gas here since the gas is being compressed.
Hence the total work done by the gas during this two step process is :
[tex]W=W_1+W_2[/tex]
[tex]W=0-3.2\times 10^4[/tex]
[tex]W=-3.2\times 10^4\ J[/tex] is the work done by the gas.