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dsp73
3 years ago
9

A 5 newton force and a 7 newton force act concurrently on a point. As the angle between the forces is increased from 0 to 180 th

e magnitude of the resultant of the two forces changes from
Physics
1 answer:
Reika [66]3 years ago
4 0

Answer:

The magnitude of the resultant decreases from A+B to A-B

Explanation:

The magnitude of the resultant of two vectors is given by

R=\sqrt{A^2 +B^2 +2AB cos \theta}

where

A is the magnitude of the first vector

B is the magnitude of the second vector

\theta is the angle between the directions of the two vectors

In the formula, A and B are constant, so the behaviour depends only on the function cos \theta. The value of cos \theta are:

- 1 (maximum) when the angle is 0, so the magnitude of the resultant in this case is

R=\sqrt{A^2 +B^2+2AB}=\sqrt{(A+B)^2}=A+B

- then it decreases, until it becomes 0 when the angle is 90 degrees, where the magnitude of the resultant is

R=\sqrt{A^2 +B^2+0}=\sqrt{A^2+B^2}

- then it becomes negative, and continues to decrease, until it reaches a value of -1 when the angle is 180 degrees, and the magnitude of the resultant is

R=\sqrt{A^2 +B^2-2AB}=\sqrt{(A-B)^2}=A-B


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Goshia [24]

Answer:

a) R_s = 0.092

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Explanation:

given,

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R_s=\left | \dfrac{n_1cos\theta_i-n_2cos\theta_t}{n_1cos\theta_i+n_2cos\theta_t} \right |^2\\R_p=\left | \dfrac{n_1cos\theta_t-n_2cos\theta_i}{n_1cos\theta_t+n_2cos\theta_i} \right |^2

using snell's law to calculate θ t

sin \theta_t = \dfrac{n_1sin\theta_i}{n_2}=\dfrac{sin45^0}{1.5}=\dfrac{\sqrt{2}}{3}

cos \theta_t =\sqrt{1-sin^2\theta_t} = \dfrac{sqrt{7}}{3}

a) R_s=\left | \dfrac{\dfrac{1}{\sqrt{2}}-\dfrac{1.5\sqrt{7}}{3}}{\dfrac{1}{\sqrt{2}}+\dfrac{1.5\sqrt{7}}{3}} \right |^2

R_s = 0.092

b) R_p=\left | \dfrac{\dfrac{\sqrt{7}}{3}-\dfrac{1.5}{\sqrt{2}}}{\dfrac{\sqrt{7}}{3}+\dfrac{1.5}{\sqrt{2}}} \right |^2

R_p = 0.085

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3 years ago
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