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valentina_108 [34]
3 years ago
12

Which of these values is equivalent to the change in momentum of

Physics
2 answers:
sasho [114]3 years ago
6 0

Answer : The correct option is, (A) impulse

Explanation :

Change in momentum : It is the product of the mass of an object and the change in velocity of an object.

Formula : \Delta p=m\times \Delta V

Impulse : It is the product of the force applied on the object and the time. It is also equivalent to the change in momentum of  an object.

Formula : J=F\times t

And the impulse is also equal to the change in momentum.

Proof of J=\Delta p :

J=F\times t\\\\J=(m\times a)\times t\\\\J=m\times (a\times t)\\\\J=m\times \Delta v=\Delta p

Force : It is the product of mass of an object and acceleration of an object.

Velocity : It is defined as the change in the position of an object in unit time.

Acceleration : It is defined as the change in the velocity of an object in a unit time.

Hence, the impulse is equivalent to the change in momentum of  an object.

iragen [17]3 years ago
3 0
<span>A. An impulse of a force changes the momentum of a body and has the same units and dimensions as momentum.</span>
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Answer:

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

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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}

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Answer:

2849.98 J

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From the question,

Work done by the boy = change Potential energy of the boy + change in kinetic energy of the boy

W = ΔP + ΔK..................... Equation 1

Where W = work done by the boy, ΔP = change in potential energy of the boy, ΔK = Change in kinetic energy of the boy.

But,

ΔP = mgΔh.................... Equation 2

ΔK = 1/2mΔv²................. Equation 3

Where m = mass of the boy, Δh = change in height of the boy, Δv = change in velocity of the boy.

Substitute equation 2 and 3 into equation 1

W = mgΔh+1/2mΔv²................. Equation 4

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