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Zepler [3.9K]
3 years ago
6

HELP PLEASECan someone please explain to me why the answer would be 2.45m/s?

Physics
2 answers:
trapecia [35]3 years ago
8 0

Solution with explanation is given below in attachment.

marissa [1.9K]3 years ago
4 0
It's basically asking for the acceleration. When an object is in free fall, the acceleration is just gravity. so in this case ∆v = 2.45 because it is the gravity on that planet.
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murzikaleks [220]

Answer:

160 Nm

Explanation:

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State Archimedes' principle
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Explanation:

Archimedes' principle states that the upward buoyant force which is exerted on body when immersed  whether fully submerged or partially in the fluid is equal to weight of fluid which body displaces and this force acts in upward direction at center of mass of displaced fluid.

Thus,

<u>Weight of the displaced fluid = Weight of the object - Weight of object in fluid.</u>

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Calculate the work efecttuated, when the force is 300 N, and the distance is 300 meters.
Alexxx [7]

Hello!

Use the formula:

W = Fd

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8 0
3 years ago
Read 2 more answers
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
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