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3 years ago

A nonzero net force acts on a particle and does work. Which one of the following statements is true?

Physics

1 answer:

arlik [135]3 years ago

5 0

Answer:

option (D) is correct.

Explanation:

According to the work energy theorem, the work done by all forces is equal to the change in kinetic energy of the body.

the kinetic energy of a body is directly proportional to the square of the speed of the body.

As the kinetic energy change, the speed of the body also change.

Option (D) is correct.

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A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

6 0

3 years ago

What is an elastic collision?

Anit [1.1K]

Answer:

A collision in which both total momentum and total kinetic energy are conserved

Explanation:

In classical physics, we have two types of collisions:

- Elastic collision: elastic collision is a collision in which both the total momentum of the objects involved and the total kinetic energy of the objects involved are conserved

- Inelastic collision: in an inelastic collision, the total momentum of the objects involved is conserved, while the total kinetic energy is not. In this type of collisions, part of the total kinetic energy is converted into heat or other forms of energy due to the presence of frictional forces. When the objects stick together after the collision, the collisions is called 'perfectly inelastic collision'

6 0

3 years ago

A point charge Q of mass M is located initially at the center of the ring. When it is displaced slightly, the point charge accel

dolphi86 [110]

Answer:

n musical notation, stems are the, "thin, vertical lines that are directly connected to the [note] head." Stems may point up or down. ... There is an exception to this rule: if a chord contains a second, the stem runs between the two notes with the higher being placed on the right of the stem and the lower on the left.

Explanation:

6 0

3 years ago

When you whisper you produce a 10-dB sound.

masha68 [24]

All you would do is for a, 10 times 2 is 20 so it would be 20-dB
For b, 10 times 4 is 40 so it would be 40-dB
<span>For c, 10 times 8 is 80 so it would be 80-dB</span>

4 0

3 years ago

The force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of

motikmotik

Explanation:

It is given that, the force needed to keep a car from skidding on a curve varies inversely as the radius of the curve and jointly as the weight of the car and the square of the car's speed such that,

$F\propto \dfrac{mgv^2}{r}$

$F=\dfrac{kmgv^2}{r}$

mg is the weight of the car

r is the radius of the curve

v is the speed of the car

Case 1.

F = 640 pounds

Weight of the car, W = mg = 2600 pound

Radius of the curve, r = 650 ft

Speed of the car, v = 40 mph

$640=\dfrac{k(2600)(40)^2}{650}$

k = 0.1

Case 2.

Radius of the curve, r = 750 ft

Speed of the car, v = 30 mph

$F=\dfrac{0.1\times 2600\times (30)^2}{750}$

F = 312 N

Hence, this is the required solution.

6 0

3 years ago