A student increased the resister of the voltmeter year not the circuit

When placed in a lake, an object either floats on the surface or sinks. It does not float at some intermediate location between

STALIN [3.7K]

The particle in the air match the mass with the particles around the balloon stoping it from floating any higher

3 0

3 years ago

Read 2 more answers

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

Estimate how long a 2500 W electric kettle would take to boil away 1.5 Kg of water . The specific latent heat of vaporization of

andrew-mc [135]

The time it would take a 2500 W electric kettle to boil away 1.5 Kg of water is 2400 seconds

<h3>How to calculate the time</h3>

Use the formula:

Power × time = mass × specific heat

Given mass = 1. 5kg

Specific latent heat of vaporization = 4000000 J/ Kg

Power = 2500 W

Substitute the values into the formula

Power × time = mass × specific heat

2500 × time = 1. 5 × 4000000

Make 'time' the subject

time = 1. 5 × 4000000 ÷ 2500 = 6000000 ÷ 2500 = 2400 seconds

Therefore, the time it would take a 2500 W electric kettle to boil away 1.5 Kg of water is 2400 seconds.

Learn more about specific latent heat of vaporization:

https://brainly.in/question/1580957

#SPJ1

8 0

2 years ago

How much momentum does a 10 kg object have when is moving 30 m/s

nikklg [1K]

P = m x v

P = 30 x 10
=300

3 0

3 years ago

Your friend is wearing a red coat. When white light hits the coat, some light is reflected, and some is absorbed.

Vilka [71]

Answer:

Orange , yellow, green and blue

red coat absorbs all colors of visible light except red, so red light

is the only light left to bounce off of the coat toward our eyes.

4 0

2 years ago