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pychu [463]
2 years ago
15

An object falls from a height of 12m. The object's potential energy at the top is 50J. Where is the potential energy of the obje

ct
25J
Physics
1 answer:
alex41 [277]2 years ago
5 0

Answer:

the height of the is 6 m.

Explanation:

Given;

potential energy of the object, P.E = 50 J

Height of fall of the object, H = 12 m

Determine the mass of the object, m

P.E = mgh

m = (P.E) / (gh)

m = (50) / (9.8 x 12)

m = 0.425 kg

Let the height of the object at 25 J = h₂

mgh₂ = 25

h₂ = (25) / (mg)

h₂ = (25) / (0.425 x 9.8)

h₂ = 6 m

Therefore, the height of the object at 25 J is 6 m.

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A 0.500-nm x-ray photon is deected through 134 in a Compton scattering event. At what angle (with respect to the incident beam)
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Answer:

The angle of recoil electron with respect to incident beam of photon is 22.90°.

Explanation:

Compton Scattering is the process of scattering of X-rays by a charge particle like electron.

The angle of the recoiling electron with respect to the incident beam is determine by the relation :

\cot\phi = (1+\frac{hf}{m_{e}c^{2}  })\tan\frac{\theta }{2}      ....(1)

Here ∅ is angle of recoil electron, θ is the scattered angle, h is Planck's constant, m_{e} is mass of electron, c is speed of light and f is the frequency of the x-ray photon.

We know that, f = c/λ      ......(2)

Here λ is wavelength of x-ray photon.

Rearrange equation (1) with the help of equation (1) in terms of  λ .

\cot\phi = (1+\frac{h}{m_{e}c\lambda  })\tan\frac{\theta }{2}

Substitute 6.6 x 10⁻³⁴ m² kg s⁻¹ for h, 9.1 x 10⁻³¹ kg for m_{e}, 3 x 10⁸ m/s for c, 0.500 x 10⁻⁹ m for λ  and 134° for θ in the above equation.

\cot\phi = (1+\frac{6.6\times10^{-34} }{9.1\times10^{-31}\times3\times10^{8}\times0.5\times10^{-9}  })\tan\frac{134 }{2}

\cot\phi=2.37

\phi = 22.90°

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