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3 years ago

A 0.2-kg ball moving at 2.0 m/s perpendicular to a wall rebounds from the wall at 1.5 m/s. The impulse given to the ball is:

Physics

1 answer:

kiruha [24]3 years ago

8 0

Answer:

J = 0.7 N-s away from wall

Explanation:

Given that,

Mass of the ball, m = 0.2 kg

Initial speed of the ball, u = 2 m/s

Final speed of the ball, v = -1.5 m/s (as it rebounds)

Impulse is equal to the change in momentum. So,

$J=m(v-u)\\\\J=0.2\times (-1.5-2)\\\\J=-0.7\ N-m$

So, the impulse given to the ball is 0.7 N-m and it is away from the wall. Hence, the correct option is (c).

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3 years ago

n alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.94 cm in a uniform magnetic field with B = 1.10 T.

9966 [12]

Answer:

a). V = 3.13*10⁶ m/s

b). T = 1.19*10^-7s

c). K.E = 2.04*10⁵

d). V = 1.02*10⁵V

Explanation:

q = +2e

M = 4.0u

r = 5.94cm = 0.0594m

B = 1.10T

1u = 1.67 * 10^-27kg

M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg

a). Centripetal force = magnetic force

Mv / r = qB

V = qBr / m

V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27

V = 2.09088 * 10^-20 / 6.68 * 10^-27

V = 3.13*10⁶ m/s

b). Period of revolution.

T = 2Πr / v

T = (2*π*0.0594) / 3.13*10⁶

T = 1.19*10⁻⁷s

c). kinetic energy = ½mv²

K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²

K.E = 3.27*10^-14J

1ev = 1.60*10^-19J

xeV = 3.27*10^-14J

X = 2.04*10⁵eV

K.E = 2.04*10⁵eV

d). K.E = qV

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V = 2.04*10⁵ / (2eV).....2e-

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3 years ago

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.

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4 years ago

What does deposition mean in a sedimentary formation

Zinaida [17]

Answer:

Sedimentary rocks are types of rock that are formed by the deposition and subsequent cementation of that material at the Earth's surface and within bodies of water. Deposition means that all the sediments, soil, and rocks are all compressed (tightly pressed into each other) and create sedimentary rocks.

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3 years ago