Ask question

Ask question

All categories

3 years ago

6

5. A 5.0 kg object accelerates uniformly from rest for 5.0 s and reaches a final velocity of 20.0 m/s. At 3.0 s, what is the obj

ect's momentum?

1 answer:

irina [24]3 years ago

3 0

Answer:

jack JACKKK

Explanation:

OK YAHHHHH SO DO THIS

You might be interested in

A battery has a potential resistance of 12 V and a current of 1280 mA. What is the resistance? PLS HELP ME ASAPPPPPPPPPP

Romashka-Z-Leto [24]

Resistance (R) = <u>9.375 ohm (Ω)</u>

Steps:

$R = \frac{V}{I}$

$= \frac{12 \: volt}{1280 \: milliampere}$

$= \frac{12 \: volt}{1.28 \: ampare}$

$= 9.375 \: ohm (Ω)$

6 0

3 years ago

A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl

Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the 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, $\theta=157^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.$

Therfore,

$\lambda^{'}-\lambda=4.64 p.m.$

Here, the photon's incident wavelength is $\lamda=7.33pm$

So,

$\lambda^{'}=7.33+4.64=11.97 p.m$

From the conservation of momentum,

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

here, $\vec{P_\lambda}$ 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,

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

Therfore,

$\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.$

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

8 0

3 years ago

.) WHat is the force that causes the pendulum to fall?<br>

soldi70 [24.7K]

Answer:

The two forces that affect on the pendulum are the force of gravity,

Explanation:

⇒ There is the force of gravity that acts downward upon the bob. It results from the Earth's mass attracting the mass of the bob. And there is a tension force acting upward and towards the pivot point of the pendulum.

Therefore, I hope this helps!

Answered by: $DreamzWorldz\\$

4 0

3 years ago

Which of the following is a correct formula of speed

satela [25.4K]

Answer:s= d/t

Explanation:

7 0

3 years ago

A freezer has a coefficient of performance of 4.1. You place 0.45 kg of water at 16°C in the freezer, which maintains its temper

velikii [3]

Ohh this is quite a lot umm i’ll be back hold on

5 0

3 years ago