Answer:
a) h=3.16 m, b) v_{cm }^ = 6.43 m / s
Explanation:
a) For this exercise we can use the conservation of mechanical energy
Starting point. Highest on the hill
Em₀ = U = mg h
final point. Lowest point
[tex]Em_{f}[/tex] = K
Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere
K = ½ m [tex]v_{cm }^{2}[/tex] + ½ [tex]I_{cm}[/tex] w²
angular and linear speed are related
v = w r
w = v / r
K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²
Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)
as there are no friction losses, mechanical energy is conserved
Em₀ = Em_{f}
mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)
h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)
for the moment of inertia of a basketball we can approximate it to a spherical shell
I_{cm} = ⅔ m r²
we substitute
h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)
h = ½ v_{cm }^{2}/g 5/3
h = 5/6 v_{cm }^{2} / g
let's calculate
h = 5/6 6.1 2 / 9.8
h = 3.16 m
b) this part of the exercise we solve the speed of equation 1
v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)
in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia
I_{cm} = ½ m r²
we substitute
v_{cm } = √ [2gh / (1 + ½)]
v_{cm } = √(4/3 gh)
let's calculate
v_{cm } = √ (4/3 9.8 3.16)
v_{cm }^ = 6.43 m / s
Albert Einstein won the Nobel Prize in Physics for his explanation, in 1905, of the photoelectric effect. Einstein showed that the results of the experiment can only be explained in terms of a particle model of light - light must be acting as if it is made up of particles (known as photons) rather than waves. This was especially fascinating since many other experiments can only be explained in terms of light acting as a wave. (a) What are the predictions of the particle theory regarding this experiment
Airplanes are launched from aircraft carriers by means of a steam catapult. The catapult is a well-insulated cylinder that contains steam, and is fitted with a frictionless piston. The piston is connected to the airplane by a cable. As the steam expands, the movement of the piston causes movement of the plane. A catapult design calls for 270 kg of steam at 15 MPa and 450°C to be expanded to 0.4 MPa. How much work can this catapult generate during a single stroke? Compare this to the energy required to accelerate a 30,000 kg aircraft from rest to 350 km per hour.
Find the screenshots in the attachment for complete solution. Follow the sequence
The catapult can generate 216,900,000 J of work during a single stroke, which is significantly more energy than the 142,222,500 J required to accelerate the aircraft.
To calculate the work done during the expansion of the steam in the catapult, we can use the first law of thermodynamics, which is the energy balance equation for a closed system:
ΔU = Q - W
Where:
ΔU is the change in internal energy of the system.
Q is the heat added to the system.
W is the work done by the system.
In this case, the steam in the catapult is the system, and we want to find the work done (W) during the expansion.
We can calculate the change in internal energy (ΔU) using the following equation:
ΔU = m * (u_final - u_initial)
Where:
m is the mass of the steam.
u_final is the specific internal energy of the steam at the final state.
u_initial is the specific internal energy of the steam at the initial state.
Given that the steam expands from 15 MPa and 450°C to 0.4 MPa, we need to find the specific internal energies at these two states. You'll typically need steam tables or thermodynamic software for precise values, but I'll provide approximate values for illustration:
For the initial state (15 MPa and 450°C), the specific internal energy might be approximately u_initial = 3270 kJ/kg.
For the final state (0.4 MPa), the specific internal energy might be approximately u_final = 2480 kJ/kg.
The mass (m) of the steam is given as 270 kg.
Now, calculate the change in internal energy:
ΔU = 270 kg * (2480 kJ/kg - 3270 kJ/kg)
ΔU = -216,900 kJ
The negative sign indicates a decrease in internal energy during the expansion.
Now, let's calculate the work done (W). Since this is an expansion, the work done by the system is positive:
W = -ΔU
W = -(-216,900 kJ)
W = 216,900 kJ
Now, let's compare this to the energy required to accelerate a 30,000 kg aircraft from rest to 350 km per hour. To calculate this energy, we can use the kinetic energy formula:
KE = 0.5 * m * v^2
Where:
KE is the kinetic energy.
m is the mass of the aircraft.
v is the velocity.
Converting 350 km per hour to m/s:
350 km/h * (1000 m/km) / (3600 s/h) ≈ 97.22 m/s
Now, calculate the kinetic energy:
KE = 0.5 * 30,000 kg * (97.22 m/s)^2
KE ≈ 142,222,500 J (joules)
So, the energy required to accelerate the aircraft is approximately 142,222,500 joules.
Comparing the work done by the catapult (216,900 kJ or 216,900,000 J) to the energy required to accelerate the aircraft (142,222,500 J), we can see that the catapult generates significantly more energy during a single stroke than is required to accelerate the aircraft. The catapult can generate much more work due to the expansion of steam, making it suitable for launching an aircraft.
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The strengths of the fields in the velocity selector of a Bainbridge mass spectrometer are B=0.500 T and E=1.2x105 V/m. The strength of the magnetic field that seperates the ions is Bo=0.750 T. A stream of single charged Li ions is found to bend in a circular arc of radius 2.32 cm. What is the mass of the Li ions?
Answer:
[tex]m = 1.16 \times 10^{-26}\ Kg[/tex]
Explanation:
Given,
Magnetic field, B = 0.5 T
Electric field, E = 1.2 x 10⁵ V/m
strength of the magnetic field that separates the ions, Bo=0.750 T
Radius, r = 2.32 cm
Relation of charge to mass ratio is given by
[tex]\dfrac{q}{m}=\dfrac{E}{BB_0R}[/tex]
[tex]m=\dfrac{qBB_0R}{E}[/tex]
Substituting all the values
[tex]m=\dfrac{1.6\times 10^{-19}\times 0.5\times 0.75\times 02.0232}{1.2\times 10^5}[/tex]
[tex]m = 1.16 \times 10^{-26}\ Kg[/tex]
Mass of Li ions is equal to [tex]m = 1.16 \times 10^{-26}\ Kg[/tex]
The mass of the Li ions is approximately [tex]\( 1.64 \times 10^{-26} \)[/tex] kg.
To find the mass of the Li ions, we can use the formula for the radius of the circular path of a charged particle moving in a magnetic field:
[tex]\[ r = \frac{m \cdot v}{q \cdot B} \][/tex]
Where:
- r is the radius of the circular path
- m is the mass of the ion
- v is the velocity of the ion
- q is the charge of the ion
- B is the magnetic field strength
We can also use the formula for the force experienced by a charged particle moving in both electric and magnetic fields:
[tex]\[ F = q \cdot (E + v \times B) \][/tex]
Since the particle moves in a circular path, the electric force F must be equal to the magnetic force [tex]\( q \cdot v \cdot B \)[/tex], where v is the speed of the particle.
We can rearrange the formulas to solve for m:
[tex]\[ m = \frac{q \cdot r \cdot B}{v} \][/tex]
Now, we need to find \( v \). Since the particle passes through both the electric and magnetic fields without deviation, the forces acting on it must be balanced. Therefore:
[tex]\[ F_{electric} = F_{magnetic} \]\[ q \cdot E = q \cdot v \cdot B \][/tex]
Solving for v:
[tex]\[ v = \frac{E}{B} \][/tex]
Now, substituting v into the equation for m :
[tex]\[ m = \frac{q \cdot r \cdot B}{\frac{E}{B}} \]\[ m = \frac{q \cdot r \cdot B^2}{E} \][/tex]
Now we can calculate \( m \) using the given values:
[tex]\[ m = \frac{(1.6 \times 10^{-19} C) \times (0.0232 m) \times (0.75 T)^2}{1.2 \times 10^5 V/m} \]\[ m= 1.64 \times 10^{-26} kg \][/tex]
Gasoline flows in a long, underground pipeline at a constant temperature of 15o C (rho = 680 kg/m3 ; ν = 4.6 × 10-7 m2 /s). Two pumping stations are located at the same elevation 13 km apart. The pressure drop between the stations is 1.4 MPa. The diameter of the pipe is 0.6 m and has a roughness of ε = 0.12 mm. Determine the volume flow rate.
Answer:
1.0416 m∧3/sec
Explanation:
check the pictures below for the solution
Write equations for both the electric and magnetic fields for an electromagnetic wave in the red part of the visible spectrum that has a wavelength of 700 nm and a peak electric field magnitude of 3.5 V/m. (Use the following as necessary: t and x. Assume that E is in volts per meter, B is in teslas, t is in seconds, and x is in meters. Do not include units in your answer. Assume that E = 0 and B = 0 when x = 0 and t = 0.) E(x, t) = B(x, t) =
Answer:
[tex]E=3.5(8.98*10^{6}x-2.69*10^{15}t)[/tex]
[tex]B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)[/tex]
Explanation:
The electric field equation of a electromagnetic wave is given by:
[tex]E=E_{max}(kx-\omega t)[/tex] (1)
E(max) is the maximun value of E, it means the amplitude of the wave.k is the wave numberω is the angular frequencyWe know that the wave length is λ = 700 nm and the peak electric field magnitude of 3.5 V/m, this value is correspond a E(max).
By definition:
[tex]k=\frac{2\pi}{\lambda}[/tex]
[tex]k=8.98*10^{6} [rad/m][/tex]
And the relation between λ and f is:
[tex]c=\lambda f[/tex]
[tex]f=\frac{c}{\lambda}[/tex]
[tex]f=\frac{3*10^{8}}{700*10^{-9}}[/tex]
[tex]f=4.28*10^{14}[/tex]
The angular frequency equation is:
[tex]\omega=2\pi f[/tex]
[tex]\omega=2\pi*4.28*10^{14}[/tex]
[tex]\omega=2.69*10^{15} [rad/s][/tex]
Therefore, the E equation, suing (1), will be:
[tex]E=3.5(8.98*10^{6}x-2.69*10^{15}t)[/tex] (2)
For the magnetic field we have the next equation:
[tex]B=B_{max}(kx-\omega t)[/tex] (3)
It is the same as E. Here we just need to find B(max).
We can use this equation:
[tex]E_{max}=cB_{max}[/tex]
[tex]B_{max}=\frac{E_{max}}{c}=\frac{3.5}{3*10^{8}}[/tex]
[tex]B_{max}=1.17*10^{-8}T[/tex]
Putting this in (3), finally we will have:
[tex]B=1.17*10^{-8}(8.98*10^{6}x-2.69*10^{15}t)[/tex] (4)
I hope it helps you!
Final answer:
To find the equations of the fields for an electromagnetic wave in the red spectrum with given parameters, calculate the frequency using the speed of light and wavelength, then apply it to the sinusoidal equations representing the electric and magnetic fields.
Explanation:
To determine the equations for the electric and magnetic fields (E and B, respectively) of an electromagnetic wave in the visible red spectrum with a wavelength (λ) of 700 nm and a peak electric field magnitude (Ē) of 3.5 V/m, we first need to calculate the frequency (f) of the wave.
Since the speed of light (c) is approximately 3 × 108 m/s, the frequency can be calculated by using the relationship c = f × λ. After finding the frequency, we can then write down the equation for the electric field E(x, t). Assuming that the wave is propagating in the +x direction and that the electric field oscillates in the y-direction, with no initial phase change, the equation for the electric field is:
E(x, t) = Ē × sin(2π(ft - x/λ))
To find the associated magnetic field B, we use the fact that the magnitudes of E and B are related by the speed of light c, such that B = E/c. Since electromagnetic waves have their electric and magnetic fields perpendicular to each other, if E is in the y-direction, B will be in the z-direction. The equation for the magnetic field is:
B(x, t) = B × sin(2π(ft - x/λ))
Where B is the peak magnetic field strength.
The radius RH of a black hole, also known as the event horizon, marks the location where not even light itself can escape from the black hole. That is, no information about the interior of the black hole may escape to any observer located outside of the black hole. According to general relativity, RH = (2GM / c^2), where M is the mass of the black hole and c is the speed of light. you want to study a black hole by getting near it ith a radial distance of 50 RH. However, you don't want the difference in gravitational acceleration between your head and your feet to exceed 10 m/s^2.
a) As a multiple of the Sun's mass, approximate what is the limit to the mass of the black hole you can tolerate at the given radial distance.
b) Is the limit an upper limit(you can tolerate smaller masses) or a lower limit (you can tolerate larger masses)?
Solution:
a) We know acceleration due to gravity, g = GM/r²
Differential change, dg/dr = -2GM/r³
Here, r = 50*Rh = 50*2GM/c² = 100GM/c ²
My height, h=dr = 1.7 m
Difference in gravitational acceleration between my head and my feet, dg = -10 m/s²
or, dg/dr = -10/1.7 = -2GM/(100GM/c²)³
or, 5.9*100³*G²*M² = 2c⁶
or, M = 0.59*c³/(1000G) = 2.39*1032 kg = [(2.39*1032)/(1.99*1030 )]Ms = 120*Ms
Mass of black hole which we can tolerate at the given distance is 120 time the mass of Sun.
b) This limit an upper limit ,we can tolerate smaller masses only.
7. A pendulum of mass 5.0 kg hangs in equilibrium. A frustrated student walks up to it and kicks the bob with a horizontal force of 30.0 ???? applied over 0.30 seconds. What is the length of the pendulum if it has a period of 5.0 seconds? What is the maximum angle of displacement of the swinging pendulum?
Answer:
Explanation:
For pendulum , time period is
T = 2π √ l / g
T² = 4π² l/ g
l = / g T² / 4π²
= 9.8 x 5² / 4π²
= 6.21 m
Angular impulse = Torque x time
= 30 x 6.21 x .3 = increase in angular momentum
30 x 6.21 x .3 = I ω ; I is moment of inertia and ω is angular velocity .
I ω = 55.89
I = 1/3 m l²
= 1/3 x 5 x 6.21²
= 64.27
Angular kinetic energy = 1/2 I ω²
= 1/2(I ω)²/ I
= .5x 55.89²/ 64.27
= 24.30 J
Due to it , the centre of mass of pendulum increases by height h so that
h = l/2 ( 1 - cos θ )
1/2 I ω² = mg l/2 ( 1 - cos θ)
24.3 = 5 x 9.8 x 3.1 ( 1 - cos θ)
( 1 - cos θ) = .16
cos θ = .84
θ = 33 degree.
The length of the pendulum is about 6.25m and the maximum angle of displacement would occur when the pendulum is kicked with the maximum force of 30N.
Explanation:The length of a pendulum is determined by the formula T=2π√(L/g). Given that the period T is 5.0 seconds and g (acceleration due to gravity) is 9.8 m/s², we can rearrange the equation to solve for L (length), resulting in L = (T² * g) / (4π²) = (5.0)² * 9.8 / (4π²) ≈ 6.25m. This regard to the maximum angle of displacement, when a pendulum is in simple harmonic motion, which is typically for displacements less than 15 degrees, the displacement is directly proportional to its force. Therefore, the maximum angle would occur when the force is the greatest, in this case, when the student kicks the pendulum with a force of 30 N.
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you and your roommates are studying hard for you physics exam. you study late into the night and then fall into your bed for some sleep. you all wake early before the exam and scramble groggily around making breakfast. you can’t agree on what to have, so one of you cooks waffles on a 990 w waffle iron while another toasts bread in a 900 w toaster. you want to make a cup of coffee with a 650 w coffeemaker, and you plug it into the same power strip into which the waffle iron and toaster are plugged. will the 20 a circuit breaker trip? explain.
Answer:
The circuit breaker will trip because the current drawn is in excess of its capacity
Explanation:
The 20 A circuit breaker is designed to not carry current in excess of its capacity.
Total power demand on power strip = 900 + 990 + 650 = 2540 w
Let us assume a standard American voltage outlet of 120 V
Recall that electric power is
P = I x V
Where I is current, and
V = voltage.
For the 2540 W power drawn, the current I is
I = P/V = 2540/120 = 21.16 A
This is 1.16 A in excess of its capacity.
A charge of 9.0 x 10-5C is placed in an electric field with a strength of 4.0 x 104 N. What is the electrical force acting
on the charge?
Answer:
3.6 N
Explanation:
E=F/q
so
F=qxE
F= 9x10^-5x4x10^4
F= 3.6 N
A 0.0140-kg bullet is fired straight up at a falling wooden block that has a mass of 2.42 kg. The bullet has a speed of 555 m/s when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time t when the collision with the bullet occurs. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time t.
Answer:
0.16 s
Explanation:
• Falling from rest (V_block= 0 m/s) the block attains a final velocity V_block before colliding with the bullet. This velocity is given by Equation 2.4 as
V_block(final velocity of block just before hitting) =V_0,block +at
where a is the acceleration due to gravity (a = —9.8 m/s2) and t is the time of fall. The upward direction is assumed to be positive. Therefore, the final velocity of the falling block is
V_block = at
• During the collision with the bullet, the total linear momentum of the bullet/block system is conserved, so we have that
(M_bullet+M_block)V_f = M_bullet*V_bullet+ M_block*V_block
Total linear momentum after collision = Total linear momentum before collision
Here V_f is the final velocity of the bullet/block system after the collision, and V_bullet and V_block are the initial velocities of the bullet and block just before the collision. We note that the bullet/block system reverses direction, rises, and comes to a momentary halt at the top of the building. This means that V_f, the final velocity of the bullet/block system after the collision must have the same magnitude as V_block, the velocity of the falling block just before the bullet hits it. Since the two velocities have opposite directions, it follows that of V_f =-V_block, Substituting this relation and Equation (1) into Equation (2) gives
(M_bullet + M_block)(-at) = M_bullet*V_bullet + M_block(at)
t = -M_bullet*V_bullet/a(M_bullet +2M_block)
=-(0.0140-kg)*555 m/s/-9.8(0.0140-kg+2(2.42 kg)
=0.16 s
Final answer:
The student's physics question involves applying the conservation of momentum and kinematic equations to determine the time the block was falling before being struck by a bullet. By considering the bullet's speed, the mass of both the bullet and block, and the motions involved, one can calculate the time of fall.
Explanation:
The student's question involves a physics problem related to conservation of momentum and projectile motion. The bullet's speed when it strikes the block, and the subsequent motion of the block and bullet system provide key information. To solve for the time t that the block was falling, we need to use the principles of physics that dictate how objects move under the influence of gravity and how they interact in collisions.
To find the time t, we'll take the following steps:
Use the conservation of momentum to find the velocity of the block and bullet immediately after the collision.
Use kinematic equations to relate this velocity to the maximum height reached by the block and bullet, taking into account the direction reversal.
Use the kinematic equations again to find the time t during which the block was falling before the collision.
Assuming the collision is perfectly inelastic, the bullet embeds itself in the block, so we have a combined mass moving upward after the collision. This combined mass will move up to the original height due to the conservation of energy principle, as the initial kinetic energy is converted entirely to gravitational potential energy at the peak of the ascent after which it momentarily comes to a halt.
The sun continuously radiates energy into space in all directions. Some of the sun's energy is intercepted by the Earth. The average temperature of the surface of the Earth remains a little above 300 k. Why doesn't the Earth's temperature rise as it intercepts the sun's energy?
A. The Earth reflects the sun's light.
B. The Earth radiates an amount of energy into space equal to the amount it receives.
C. The energy only raises the temperature of the upper atmosphere and never reaches the surface.
D. The thermal conductivity of the Earth is low.
E. The heat is carried away from the Earth by convection currents.
B. The Earth radiates an amount of energy into space equal to the amount it receives.
Part of the solar energy is reflected by the Earth into space, this is known as albedo. The other part of the energy radiated by the Earth in the form of infrared radiation, is absorbed by the greenhouse gases, which cause most of this infrared radiation to be emitted into space. Therefore, the net flow of energy is zero.
1200 N-m of torque is used to drive a gear (A) of diameter 25 cm, which in turn drives another gear (B) of diameter 52 cm. What torque is transmitted by gear B to its shaft? Give your answer in kN-m.
Answer:
2.5kN.m
Explanation:
Torque is directly proportional to pitch diameter
= Ta/Tb= Da/Db
=120/Tb= 0.25/0.5
Tb= 2.469kN.m approx 2.5kN.m
Answer:
Explanation:
answer explain below
Ch 26 HW
Gedanken Conceptual Questions
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Gedanken Conceptual Questions
Einstein developed much of his understanding of relativity through the use of gedanken, or thought, experiments. In a gedanken experiment, Einstein would imagine an experiment that could not be performed because of technological limitations, and so he would perform the experiment in his head. By analyzing the results of these experiments, he was led to a deeper understanding of his theory.
In each the following gedanken experiments, Albert is in the exact center of a glass-sided freight car speeding to the right at a very high speed v relative to you.
Albert has a flashlight in each hand and directs them at the front and rear ends of the freight car. Albert switches the flashlights on at the same time.
Part A
In Albert's frame of reference, which beam of light travels at a greater speed, the one directed toward the front or the one toward the rear of the train, or do they travel at the same speed? Which beam travels faster in your frame of reference?
Enter the answers for Albert's frame of reference and your frame of reference separated by a comma using the terms front, rear, and same. For example, if in Albert's frame of reference the beam of light directed toward the front of the train travels at a greater speed and in your frame of reference the two beams travel at the same speed, then enter front,same.
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Part B
In Albert's frame of reference, which end, front or rear, is struck by light first, or are they struck at the same time? Which end is struck first in your frame of reference?
Enter the answer for Albert's frame of reference and your frame of reference separated by a comma using the terms front, rear, and same. For example, if in Albert's frame of reference the beam of light strikes the front first and in your frame of reference the two beams strike at the same time, then enter front,same.
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Now Albert directs his flashlights at the ceiling and floor of the freight car. The flashlights are located midway between the ceiling and the floor and Albert switches them on at the same time
Part C
In Albert's frame of reference, which surface, ceiling or floor, is struck by light first, or are they struck at the same time? Which surface is struck first in your frame of reference?
Enter the answer for Albert's frame of reference and your frame of reference separated by a comma using the terms ceiling, floor, and same. For example, if in Albert's frame of reference the ceiling is struck by light first and in your frame of reference the floor and the ceiling are struck by light at the same time, then enter ceiling,same.
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Albert is playing laser tag in the freight car. Two "assassins" sneak into the freight car with Albert. One is positioned against the front end and the other against the rear end. They each fire a laser at Albert. The two lasers strike Albert at the same time.
Part D
In Albert's frame of reference, who fired first, the person against the rear end or the person against the front end, or did they fire at the same time? In your frame of reference, who fired first?
Enter the answers for Albert's frame of reference and your frame of reference separated by a comma using the terms front, rear, and same. For example, if in Albert's frame of reference the person against the front end fired first and in your frame of reference both "assassins" fired at the same time, then enter front,same.
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Answer:
Please see the attached picture for the answer.
Explanation:
Final answer:
Einstein's Gedanken experiments reveal the constant speed of light and challenge intuitive notions of time and simultaneity in special relativity. In Albert's reference frame and an outside observer's frame, light beams travel at the same speed, but the perception of simultaneity can differ due to relativistic effects.
Explanation:
Gedanken Experiments and Einstein's Theory of Relativity
Albert Einstein used Gedanken's experiments to explore the implications of relativity on the concepts of time and simultaneity. These thought experiments challenged intuitive notions by demonstrating how the speed of light remains constant in all frames of reference, a cornerstone of special relativity. Let's address the student's questions based on this principle:
Part A: In both Albert's frame of reference and your frame of reference, the beams of light travel at the same speed since the speed of light is constant regardless of the observer's motion relative to the source.
Part B: Albert sees both beams of light strike the ends of the freight car simultaneously. However, due to relativistic effects, a stationary observer might see the light traveling in the direction of the freight car's motion strike first.
Part C: Both in Albert's and your frame of reference, the light strikes the ceiling and floor simultaneously because there is no relative motion along the direction of the light beams' propagation.
Part D: In Albert's frame of reference, both lasers strike him simultaneously. For a stationary observer, depending on the direction of the freight car's motion, one assassin may seem to fire before the other due to relativistic effects.
Einstein's ingenious analytical methods demonstrated through Gedanken experiments, led to the conclusion that many seemingly obvious predictions must be reconsidered when dealing with the postulates of relativity, which have been confirmed through meticulous experimentation.
A person with mass mp = 74 kg stands on a spinning platform disk with a radius of R = 2.31 m and mass md = 183 kg. The disk is initially spinning at ω = 1.8 rad/s. The person then walks 2/3 of the way toward the center of the disk (ending 0.77 m from the center). 1)What is the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk? kg-m2 2)What is the total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk? kg-m2 3)What is the final angular velocity of the disk? rad/s 4)What is the change in the total kinetic energy of the person and disk? (A positive value means the energy increased.) J 5)What is the centripetal acceleration of the person when she is at R/3? m/s2 6)If the person now walks back to the rim of the disk, what is the final angular speed of the disk? rad/s
Answer:
1) 883 kgm2
2) 532 kgm2
3) 2.99 rad/s
4) 944 J
5) 6.87 m/s2
6) 1.8 rad/s
Explanation:
1)Suppose the spinning platform disk is solid with a uniform distributed mass. Then its moments of inertia is:
[tex]I_d = m_dR^2/2 = 183*2.31^2/2 = 488 kgm^2[/tex]
If we treat the person as a point mass, then the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk:
[tex]I_{rim} = I_d + m_pR^2 = 488 + 74*2.31^2 = 883 kgm^2[/tex]
2) Similarly, he total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk (1/3 of the radius from the center):
[tex]I_{R/3} = I_d + m_p(R/3)^2 = 488 + 74*(2.31/3)^2 = 532 kgm^2[/tex]
3) Since there's no external force, we can apply the law of momentum conservation to calculate the angular velocity at R/3 from the center:
[tex]I_{rim}\omega_{rim} = I_{R/3}\omega_{R/3}[/tex]
[tex]\omega_{R/3} = \frac{I_{rim}\omega_{rim}}{I_{R/3}}[/tex]
[tex]\omega_{R/3} = \frac{883*1.8}{532} = 2.99 rad/s[/tex]
4)Kinetic energy before:
[tex]E_{rim} = I_{rim}\omega_{rim}^2/2 = 883*1.8^2/2 = 1430 J[/tex]
Kinetic energy after:
[tex]E_{R/3} = I_{R/3}\omega_{R/3}^2/2 = 532*2.99^2/2 = 2374 J[/tex]
So the change in kinetic energy is: 2374 - 1430 = 944 J
5) [tex]a_c = \omega_{R/3}^2(R/3) = 2.99^2*(2.31/3) = 6.87 m/s^2[/tex]
6) If the person now walks back to the rim of the disk, then his final angular speed would be back to the original, which is 1.8 rad/s due to conservation of angular momentum.
If the wave speed is 20 m/s and the wave length is 2 meters find the frequency ?
Answer:
10
Explanation:
v=f×lander
20=f2
f=10
STATE THE LAWS OF REFLECTION?
Answer:
hii bro.... from India. ....i am from Bilaspur cg
A flywheel of mass M is rotating about a vertical axis with angular velocity ω0. A second flywheel of mass M/5 is not rotating and drops onto the first flywheel and sticks to it. Both flywheels have radius R. What is the final angular velocity in terms of the initial angular velocity ω0? Treat each flywheel as a disk (I = (1/2) MR2).
Answer:
0.83 ω
Explanation:
mass of flywheel, m = M
initial angular velocity of the flywheel, ω = ωo
mass of another flywheel, m' = M/5
radius of both the flywheels = R
let the final angular velocity of the system is ω'
Moment of inertia of the first flywheel , I = 0.5 MR²
Moment of inertia of the second flywheel, I' = 0.5 x M/5 x R² = 0.1 MR²
use the conservation of angular momentum as no external torque is applied on the system.
I x ω = ( I + I') x ω'
0.5 x MR² x ωo = (0.5 MR² + 0.1 MR²) x ω'
0.5 x MR² x ωo = 0.6 MR² x ω'
ω' = 0.83 ω
Thus, the final angular velocity of the system of flywheels is 0.83 ω.
A man is traveling on a bicycle at 14 m/s along a straight road that runs parallel to some railroad tracks. He hears the whistle of a train that is behind him. The frequency emitted by the train is 840 Hz, but the frequency the man hears is 778 Hz. Take the velocity of sound to be 340 m/s.
a) What frequency is heard by a stationary observer located between the train and the bicycle?
b) What is the speed of the train, and is the train traveling away from or towardthe bicycle?
Answer:
Explanation:
b ) The problem is based on Doppler's effect of sound
f = f₀ x (V - v₀) /( [tex]V+v_s[/tex])
f is apparent frequency ,f₀ is real frequency , V is velocity of sound , v₀ is velocity of observer going away , [tex]v_s[/tex] is velocity of source going away
778 = 840 x (340 - 14)/ (340 + [tex]v_s[/tex])
340 + [tex]v_s[/tex] = 341.18
[tex]v_s[/tex] = 1.18 m /s
it will go away from the observer or the cyclist.
speed of train = 1.18 m /s
a )
For a stationary observer v₀ = 0
f = f₀ x V /( [tex]V+v_s[/tex])
= 840 x 340 / (340 + 1.180)
= 837 Hz
Final answer:
The stationary observer between the train and the bicycle hears a frequency of 809 Hz, and the train is moving away from the bicycle at 42 m/s.
Explanation:
A stationary observer located between the train and the bicycle would hear a frequency of 809 Hz as the train approaches and passes by.
The speed of the train can be calculated to be 42 m/s moving away from the bicycle.
You are using a hand-held sound level meter to measure the intensity level of the roars produced by a lion prowling in the high grass. To decrease the measured sound intensity level by 36.2 dB requires the lion move away from you until its distance from you has increased by what factor? g
Answer:
64.57
Explanation:
We are given that
For decreasing measured sound intensity level=36.2 dB
We have to find the factor by which the distance increase.
Let initial distance=x
Final distance=x'
According to question
[tex]36.2=10log(\frac{x'^2}{x^2})[/tex]
[tex]36.2=10log(\frac{x'}{x})^2[/tex]
[tex]36.2=10\times 2log\frac{x'}{x}[/tex]
[tex]log\frac{x'}{x}=\frac{36.2}{20}=1.81[/tex]
[tex]\frac{x'}{x}=10^{1.81}[/tex]
[tex]x'=10^{1.81}x=64.57x[/tex]
Hence, the distance is increases by factor of 64.57
An infinitely long, straight, cylindrical wire of radius R R has a uniform current density → J = J ^ z J→=Jz^ in cylindrical coordinates. What is the magnitude of the magnetic field at some point inside the wire at a distance r i < R ri
Answer:
B(2 pi r) = uo i = uo (J pi r^2)
Magnetic field, B = μo Jr/2
Explanation:
The magnitude of the magnetic field inside the wire at a distance r1 < R inside the wire is;
B = ½μ_o × Jr
We want to find the magnitude of the magnetic field inside the wire at a distance r1 < R.
The formula for Magnetic field at a distance r located inside a conductor of radius R is given by Ampere's circuital law as;
B × 2πr = μ_o × Jπr²
Where;
B is magnitude of magnetic field
μ_o is a constant known as magnetic permeability
J is current density
Thus;
B × 2πr = μ_o × Jπr²
πr will cancel out from both sides to give;
2B = μ_o × Jr
B = ½μ_o × Jr
Thus, in conclusion them magnitude of the magnetic field is ½μ_o × Jr
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A ventriloquist is able to convince you that words are coming from his dummy’s mouth because the dummy’s mouth movements match the timing of the words actually coming from the ventriloquist. The ventriloquist is making use of the perceptual principle of
The ventriloquist uses the perceptual principle known as the McGurk Effect to create the illusion that words are coming from their dummy's mouth. This involves syncing their speech with the movements of the dummy's mouth and mastering aspects of speech production. We perceive the words to be coming from the dummy due to the combined audiovisual cues.
Explanation:A ventriloquist convinces us that words are coming from their dummy's mouth by making use of the perceptual principle known as the McGurk Effect. This audiovisual speech perception demonstrates that we don't just hear speech but also 'see' it. The ventriloquist creates the illusion by carefully syncing the opening and closing of the dummy's mouth with their own speech. This overlaid speech and vision information persuades us to perceive the sound as coming from the dummy.
Human speech itself is produced by shaping the cavity formed by the throat and mouth, the vibration of vocal cords, and using the tongue to adjust the fundamental frequency and combination of those sounds. Thus, excellent ventriloquists master these aspects of speech production without visibly moving their own mouth or lips, reinforcing the illusion.
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To understand length contraction and time dilation. An inertial frame of reference is one in which Newton's laws hold. Any frame of reference that moves at a constant speed relative to an inertial frame of reference is also an inertial frame. The proper length l0 of an object is defined to be the length of the object as measured in the object's rest frame. If the length of the object is measured in any other inertial frame, moving with speed u relative to the object's rest frame (in a direction parallel to l0), the resulting length l is given by the length contraction equation:
Answer:
l = l0 ( 1 - u^2/v^2) ^0.5
Explanation:
This is the equation of lenght contraction of special relativity
An alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 4.29 cm in a uniform magnetic field with B = 1.53 T. Calculate (a) its speed, (b) its period of revolution, (c) its kinetic energy, and (d) the potential difference through which it would have to be accelerated to achieve this energy.
Answer:
A) speed = 3144287.425 m/s
B) period = 8.57x10^-8 s
C) KE = 3.302x10^-14 J
D) potential difference = 103187.5 V
Explanation:
Detailed explanation and calculation is shown in the image below.
Suppose a car approaches a hill and has an initial speed of
106
k
m
/
h
at the bottom of the hill. The driver takes her foot off the gas pedal and allows the car to coast up the hill.
A. If in actuality a
780
k
g
car with an initial speed of
106
k
m
/
h
is observed to coast up a hill and stops at a height
21.5
m
above its starting point, how much thermal energy was generated by friction in
J
?
B. What is the magnitude of the average force in newtons of friction if the hill has a slope of
27
∘
above the horizontal.
Answer:
a) 1.73*10^5 J
b) 3645 N
Explanation:
106 km/h = 106 * 1000/3600 = 29.4 m/s
If KE = PE, then
mgh = 1/2mv²
gh = 1/2v²
h = v²/2g
h = 29.4² / 2 * 9.81
h = 864.36 / 19.62
h = 44.06 m
Loss of energy = mgΔh
E = 780 * 9.81 * (44.06 - 21.5)
E = 7651.8 * 22.56
E = 172624.6 J
Thus, the amount if energy lost is 1.73*10^5 J
Work done = Force * distance
Force = work done / distance
Force = 172624.6 / (21.5/sin27°)
Force = 172624.6 / 47.36
Force = 3645 N
Two boxes, mA = 18 kg and mB = 14 kg, are attached by a string under tension T1 . The rightmost box is being pulled horizontally across the floor by a different string under tension T2 . The coefficient of kinetic friction between the boxes and the floor is µK = 0.240. If the boxes are accelerating at 3.5 m/s2 to the right, what is the tensions T1 and T2 ?
Answer:
[tex]T_{1}=105.38 N[/tex]
[tex]T_{2}=187.34 N[/tex]
Explanation:
Applying the second Newton's law for the first box we have.
[tex]-f_{1f}+T_{1}=m_{A}a[/tex]
[tex]-\mu_{k}N_{1}+T_{1}=m_{A}a[/tex]
We know that the normal force is the product between the weight and the kinetic friction, so we have:
[tex]-\mu_{k}m_{A}g+T_{1}=m_{A}a[/tex]
Now we can find T₁:
[tex]\mu_{k}m_{A}g+m_{A}a=T_{1}[/tex]
The acceleration is the same for both boxes.
[tex]T_{1}=m_{1}(\mu_{k}g+a)[/tex]
[tex]T_{1}=18*(0.240*9.81+3.5)[/tex]
[tex]T_{1}=105.38 N[/tex]
Now let's analyze the forces of the second box.
[tex]-f_{2f}-T_{1}+T_{2}=m_{B}a[/tex]
[tex]-\mu_{k}m_{B}g-T_{1}+T_{2}=m_{B}a[/tex]
Let's solve it for T₂.
[tex]T_{2}=m_{B}a+T_{1}+\mu_{k}m_{B}g[/tex]
[tex]T_{2}=m_{B}a+T_{1}+\mu_{k}m_{B}g[/tex]
[tex]T_{2}=m_{B}(a+\mu_{k}g)+T_{1}[/tex]
[tex]T_{2}=14(3.5+0.240*9.81)+105.38[/tex]
[tex]T_{2}=187.34 N[/tex]
I hope it helps you!
Suppose your telescope has perfect mirrors and is limited only by diffraction. If you are viewing with light of 400 nm wavelength, what diameter mirror in a space telescope would have 1.0-km resolution of objects on Mars when viewed from above the earth at a time when the Earth-Mars distance of 80 million kilometers
Answer:
39 m
Explanation:
We are given that
Wavelength=[tex]\lambda=400 nm=400\times 10^{-9} m[/tex]
1 nm=[tex]10^{-9} m[/tex]
y=1 km=[tex]1000 m[/tex]
1 km=1000 m
Earth mars distance =x=80 million Km=[tex]80\times 10^9 m[/tex]
1million km=[tex]10^9 m[/tex]
[tex]sin\theta=\frac{1.22\lambda}{d}[/tex]
[tex]sin\theta=\frac{y}{x}[/tex]
[tex]\frac{y}{x}=\frac{1.22\lambda}{d}[/tex]
[tex]\frac{1000}{80\times 10^9}=\frac{1.22\times 400\times 10^{-9}}{d}[/tex]
[tex]d=\frac{1.22\times 400\times 10^{-9}\times 80\times 10^9}{1000}[/tex]
[tex]d=39.04 m\approx 39 m[/tex]
Diameter of mirror =39 m
To achieve a 1.0-km resolution on Mars with light of a 400 nm wavelength, the size of the space telescope's mirror would need to be approximately 6 mm. This calculation is based on diffraction limit theory and may not be precisely accurate in practical application due to other potential limiting factors.
Explanation:This question is related to the resolution of a telescope which is primarily influenced by the aperture size and wavelength of light being observed. The bigger the telescope aperture (the diameter of the primary mirror), the greater the resolution as more light can be collected. The formula for the resolution limit due to diffraction (the smallest distinguishable detail) can be given by R = 1.22λ/D where λ is the wavelength of light and D is the diameter of the telescope's main mirror.
Here the wavelength given is 400 nm or 400 x 10^-9 m, and we need to calculate the mirror diameter required for a resolution of 1.0-km on Mars from Earth. Let's convert the resolution limit from kilometers to meters (which gives us 1000m), and then to radians (using the Earth-Mars distance), which results in an angle of 1.25x10^-11 radians approximately.
Substituting these values in the formula, we can solve for D and find that the telescope mirror size needed would be approximately 0.006 m or 6 mm for this resolution. Do note, that this is a theoretical value, in reality, the size might need to be larger due to factors like the diffraction limit, non-uniformities in mirrors, or aberrations in lenses.
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A velocity selector, in which charged particles of a specific speed pass through undeflected while those of greater or lesser speeds are deflected either to the left or to the right. After leaving the velocity selector, particles enter a purely magnetic field. A particle's radius of motion is then used to find the mass-to-charge ratio of the particle, which in turn can identify it. Taken altogether, this device is called a mass spectrometer. The illustrated electric field is directed to the right with magnitude 1.95 times 10^5 V/m and the magnetic field is directed into the screen with magnitude 0.555 T.
a. Determine the speed v of the undeflected charged particle.
b. After passing through the velocity selector, the charged particle moves in a circular path with a radius of r = 6.61 mm. Determine the particle's mass-to-charge ratio. m/q = kg/C
Answer:
a) 351351.35m/s
b) 1.044*10^{-8}kg/C
Explanation:
a) Electric force and magnetic force over the charge must have the same magnitude. From there we can compute the seep of the charge.
[tex]F_E=F_B\\\\qE=qvB\\\\v=\frac{E}{B}=\frac{1.95*10^{5}V/m}{0.555T}=351351.35\frac{m}{s}[/tex]
b) the mass-charge ratio is given by:
[tex]\frac{m}{q}=\frac{rB}{v}=\frac{(6.61*10^{-3}m)(0.555T)}{351351.35m/s}=1.044*10^{-8}\frac{kg}{C}[/tex]
hope this helps!!
A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 Joules (with the potential energy equal to zero at ground level) and is moving upward with a kinetic energy of 50 Joules. What is the maximum height h reached by the ball? Consider air friction to be negligible.
A. h ≈ 30 m
B. h ≈ 40 m
C. h ≈ 50 m
D. h ≈ 10 m
E. h ≈ 20 m
Given that,
A ball is thrown upward. At a height of 10 meters above the ground, the ball has a potential energy of 50 Joules. It is moving upward with a kinetic energy of 50 Joules.
We need to find the maximum height h reached by the ball. Let at a height of 10 meters, it has a potential energy of 50 Joules. So,
[tex]mgH=50\\\\mg=\dfrac{50}{h}\\\\mg=\dfrac{50}{10}\\\\mg=5\ N[/tex] ........(1)
Let at a height of h m, it reaches to a maximum height. at this point, it has a total of 100 J of energy. So,
[tex]mgh=50+50\\\\mgh=100\\\\h=\dfrac{100}{5}\\\\h=20\ m[/tex]
So, the correct option is (E) "h = 20 m".
The maximum height reached by the ball is about 20 meters, as determined by using the conservation of mechanical energy principle, considering that the total mechanical energy at the height of 10 meters was 100 Joules.
Explanation:To solve for the maximum height reached by the ball, we can use the conservation of mechanical energy principle, which states that the total mechanical energy (potential energy + kinetic energy) of the ball remains constant in the absence of air friction.
At 10 meters above the ground, the ball has a potential energy (PE) of 50 Joules and a kinetic energy (KE) of 50 Joules. Therefore, the total mechanical energy at that height is:
PE + KE = 50 J + 50 J = 100 J
As the ball rises, its kinetic energy is converted into potential energy until the kinetic energy becomes zero at the maximum height. The total mechanical energy at maximum height will be equal to the potential energy:
PE at maximum height = total mechanical energy = 100 J
Using the formula for gravitational potential energy, PE = mgh (where m is mass, g is the acceleration due to gravity (9.81 m/s²), and h is the height), and knowing that the PE at 10 meters is 50 J, we can find the mass of the ball:
50 J = m * 9.81 m/s² * 10 m
m = 50 J / (9.81 m/s² * 10 m) = 0.51 kg
With the mass of the ball, we can now calculate the maximum height using the total mechanical energy:
100 J = 0.51 kg * 9.81 m/s² * h
h = 100 J / (0.51 kg * 9.81 m/s²) ≈ 20 meters
Therefore, the maximum height h reached by the ball is about 20 meters.
Consider two blocks of copper. Block A contains 800 atoms and initially has a total of 20 quanta of energy. Block B contains 200 atoms and initially has 80 quanta of energy. The two blocks are placed in contact with each other, inside an insulated container (so no thermal energy can be exchanged with the surroundings). After waiting for a long time (for example, an hour), which of the following would you expect to be true?
a. Approximately 50 quanta of energy are in block A, and approximately 50 quanta of energy are in block B.
b. Approximately 80 quanta of energy are in block A, and approximately 20 quanta of energy are in block B.
c. The entropy of block A is equal to the entropy of block B.
d. The temperature of block A and the temperature of block B are equal.
Answer:
Option B and Option D are true
Explanation:
We are given;
Number of atoms in block A = 800
Energy content in block A = 20 quanta
Number of atoms in block B = 200
Energy content in block B = 80 quanta
The energy of a system which is an extensive quantity,depends on the mass or number of moles of the system. However, at equilibrium, the energy density of the two copper blocks will be equal. That is, each atom of Cu in the two blocks will, on average, have the same energy. Because block A has 4 times more atoms than block B, it will have 4 times more quanta of energy. Thus, option B is therefore true while option A is false.
Temperature is a measure of the average kinetic energy of the atoms in a material. Now, if each atom in blocks A and B have the same average energy, then the temperatures of blocks A and B will be equal at equilibrium. Thus, option D is true.
Entropy of a system is an extensive quantity that depends on the the mass or number of atoms in the system. Because block A is bigger than block B, it will have higher entropy. However, that the specific entropy (the entropy per mole or per unit mass) is an intensive quantity -- it is independent of the size of a system. The molar entropy of blocks A and B are equal at equilibrium. Thus option C is false.
For the given condition in the paragraph, the statements (a) and (c) are false. And the statements given in option (b) and (d) are true.
Given data:
Number of atoms in Block A is, 800 atoms.
Initial amount of energy in Block A is, 20 quanta.
Number of atoms in Block B is, 200 atoms.
Initial amount of energy in Block B is, 80 quanta.
We will use the concept of Energy to relate it with the temperature, to solve the given problem. Also, there is some glance about the equilibrium condition as discussed:
The energy of a system is an extensive quantity because it depends on the mass or number of moles of the system. However, at equilibrium, the energy density of the two copper blocks will be equal. That is, each atom of Cu in the two blocks will, on average, have the same energy. Because block A has 4 times more atoms than block B, it will have 4 times more quanta of energy. Thus, option B is true while option A is false.Entropy of a system is an extensive quantity that depends on the the mass or number of atoms in the system. Because block A is bigger than block B, it will have higher entropy. However, that the specific entropy (the entropy per mole or per unit mass) is an intensive quantity because it is independent of the size of a system. The molar entropy of blocks A and B are equal at equilibrium. Thus option C is false.Temperature is a measure of the average kinetic energy of the atoms in a material. If each atom in blocks A and B have the same average energy, then the temperatures of blocks A and B will be equal at equilibrium. Thus, option D is true.
Thus, we conclude that for the given condition, the statements (a) and (c) are false. And the statements given in option (b) and (d) are true.
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A uniform electric field, with a magnitude of 650 N/C, is directed parallel to the positive x-axis. If the potential at x = 3.0 m is 1 700 V, what is the potential at x = 1.0 m?
The potential at x = 1.0 m in the given uniform electric field, with the given parameters, would be 400 V.
Explanation:The subject question pertains to the concept of electric potential in a uniform electric field. The potential difference, V, in an uniform electric field is given by the equation V = Ed, where E is the field strength and d is the distance. Given an electric field strength, E, of 650 N/C and a potential, V, at x = 3.0 m of 1 700 V, we seek the potential at x = 1.0m. Since the electric field is uniform, we know that the change in potential is linear with distance. Because the distance decreases by 2 meters (from x = 3.0 m to x = 1.0 m), the potential will decrease by E*2, which is 650 N/C * 2 m = 1300 V. So, the potential at x = 1.0 m is 1 700 V - 1 300 V = 400 V.
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To find the electric potential at x = 1.0 m in a uniform electric field directed along the x-axis, subtract the potential difference due to the field over the distance moved from the initial potential; the answer is 400 V.
The question is asking to calculate the electric potential at a certain point given a uniform electric field and an initial electric potential at a different point. To find the potential at x = 1.0 m, we can use the relationship between electric field (E), electric potential (V), and distance (d), with the following formula: V = V_0 - E * d, where V is the potential at the point of interest, V_0 is the initial potential, E is the electric field strength, and d is the distance between the points.
Gor the given electric field of 650 N/C directed parallel to the positive x-axis, and with the potential at x = 3.0 m being 1700 V, the potential difference \\Delta V caused by moving from x = 3.0 m to x = 1.0 m can be calculated by multiplying the electric field strength by the distance over which the field is applied, which is 2.0 m (3.0 m - 1.0 m). Hence, \\Delta V = E * d = 650 N/C * 2.0 m = 1300 V.
Therefore, the potential at x = 1.0 m is V = 1700 V - 1300 V = 400 V.