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2 years ago

Please help only have a short amount of time left!

Physics

2 answers:

VARVARA [1.3K]2 years ago

6 0

Answer:

I think all will display the same value

Explanation:

Sorry if this is wrong

Savatey [412]2 years ago

6 0

Answer:

I think it's option a. this is because,it has longer conductivity.

this might be wrong. I am just in jss2 but I read physics

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The Trojan asteroids are found:_________ a) orbiting around the Kuiper Belt body Hector. b) beyond Neptune, with orbits similar

xenn [34]

Answer:

The correct option is;

c) sixty degrees ahead or behind Jupiter, sharing its orbit about the Sun.

Explanation:

The Trojan asteroids are the Jupiter trojans consists of asteroid that are on the same orbit as Jupiter while moving around the Sun. The Trojans can be located at the points Lagrange points L4 and L5, which are 60° ahead and 60° behind Jupiter's orbit respectively.

The first Trojan asteroid to be detected was 588 Achilles by Max Wolf in 1906. At at October, the total number of the identified Trojan asteroid was 7,040.

5 0

3 years ago

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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

Which two points on the wave shown in the diagram below are in phase with each other?

NNADVOKAT [17]

Answer:

4. B and D

Explanation:

Two points along a transverse wave (such as the one in the figure) are said to be in phase when:

- the vertical position of the two points is the same

- The oscillation of the wave is going in the same way for both points

Basically, we say that two points are in phase when they are separated by a complete cycle (one complete oscillation) of the wave.

For this wave, we see that point B and C have same displacement, but they are not in phase since in B the oscillation is going down while in C is going up.

Instead, B and D are in phase, because they are separated by one complete cycle: both points have same displacement and the oscillation is going in the same way for both of them.

8 0

3 years ago

? Which statement is true of an object in equilibrium?

Degger [83]

The answer is C,<span> The sum of all forces acting on the object is zero. hope that helps!!</span>

7 0

3 years ago

the mass of a lump of gold is constant everywhere but its weight isnt explain both in weight and mass

DiKsa [7]

That is very true, but what is the question asking you.

3 0

3 years ago

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