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saveliy_v [14]
10 months ago
5

The force diagram shown here describes the forces that external objects (the surface andEarth) exert on a woman (in this scenari

o, the force diagram does not change with time). Describe three different types of motion of the woman that are consistent with the force diagram.

Physics
1 answer:
sveticcg [70]10 months ago
4 0

The three different motions are;

  • The upward motion of the woman is constant
  • The downward motion of the woman is also constant
  • The horizontal motion of the woman is zero.

<h3>What is force diagram?</h3>

Force diagram is a pictorial or graphical illustration of different forces acting on object.

In this given question, there two forces acting on the woman as depicted in the force diagram.

  • The first force is surface force (Fs)
  • The second force is force of Earth (FE)

In the given force diagram, the woman is in equilibrium, this implies that the surface force and the Earth force are equal.

The three different types of motion of the woman that are consistent with the force diagram include the following;

  • The upward motion of the woman is constant
  • The downward motion of the woman is also constant
  • The horizontal motion of the woman is zero since there is no horizontal force on the woman.

Learn more about force diagram here: brainly.com/question/3624253

#SPJ1

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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.

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the frequency of a beam of uv light is 1.0 ×10 ^15hz what is the energy in one quantum of this light express it in ev? ​
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Answer:

4.14 eV

Explanation:

f = 1.0 ×10^15 Hz

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As 1eV = 1.6×10^-19 J so changing energy in eV from joules we will divide energy by 1.6×10^-19

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