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2 years ago

A toy remote control car drives in a circle (the controller is stuck). It makes 8 revolutions in 14

Physics

1 answer:

IceJOKER [234]2 years ago

7 0

Answer:

Period for 1 revolution is 1.75 seconds

Explanation:

given data

revolutions R = 8

time t = 14 seconds

to find out

What is the period

solution

we know that Period is the time per revolution

so here period formula that is express as

period = $\frac{R}{t}$ = $\frac{8}{14}$ = 0.57 revolution in one second

so in 1 revolution = $\frac{1}{0.57}$ seconds

so in 1 revolution = 1.75 seconds

so period for 1 revolution is 1.75 seconds

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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

How do land and sea breeze work?

Studentka2010 [4]

by the wind and air flow in the wind

7 0

3 years ago

Consider Newton's Law of Universal Gravitation: FG= G (m1 m2)/d2 .

bixtya [17]

Answer:

Mass is directly proportional to the Force of Gravity. If Mass increases, then the Force of Gravity increases; however, Distance is indirectly (or inversely) proportional to the Force of Gravity. If Distance increases, then the Force of Gravity decreases.

Explanation:

The formula for the force of gravity between two objects is

$F=G\frac{m_1 m_2}{d^2}$

where

G is the gravitational constant

m1, m2 are the masses of the two objects

d is the separation between the two objects

We notice the following:

- F is directly proportional to the masses, $F\propto m_1, m_2$ . This means that if one of the masses increases, then the force between them, F, increases in a proportional way

- F is inversely proportional to the square of the distance, $F\propto \frac{1}{d^2}$ . This means that if the distance between the two objects is increased, the force between them will decrease, and vice-versa.