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KATRIN_1 [288]
3 years ago
9

A 32-n force acts on a small mass in the positive x-direction. a 26-n force also acts on it in the negative x-direction. what is

the equilibrant of these two forces?
Physics
1 answer:
motikmotik3 years ago
6 0

Equilibrium force is the force that will keep the small mass in place, hence no movement must be made. So we know that 32 N of force is acted towards the positive direction so +32 N. Which is counteracted by 26 N force so:

32 N – 26 N = 6 N (positive)

Since positive 6 is left, therefore this must be acted by an equilibrant negative 6 N.

 

Answer:

<span>- 6 N </span>

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Two runners start at the same point on a straight track. The first runs with constant acceleration so that he covers 98 yards in
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Answer:

94.13 ft/s

Explanation:

<u>Given:</u>

  • t = time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
  • s = distance to be moved by the rock long the horizontal = 98 yards
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<u>Assume:</u>

  • u = magnitude of initial velocity of the rock
  • \theta = angle of the initial velocity with the horizontal.

For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.

\therefore y = u\sin \theta t +\dfrac{1}{2}(-g)t^2\\\Rightarrow 0 = u\sin \theta 5 +\dfrac{1}{2}(-9.8)\times 5^2\\\Rightarrow u\sin \theta 5 =\dfrac{1}{2}(9.8)\times 5^2......(1)\\

Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.

\therefore u\cos \theta t = s\\\Rightarrow u\cos \theta 5 = 98.....(2)\\

On dividing equation (1) by (2), we have

\tan \theta = \dfrac{25}{20}\\\Rightarrow \tan \theta = 1.25\\\Rightarrow \theta = \tan^{-1}1.25\\\Rightarrow \theta = 51.34^\circ

Now, putting this value in equation (2), we have

u\cos 51.34^\circ\times  5 = 98\\\Rightarrow u = \dfrac{98}{5\cos 51.34^\circ}\\\Rightarrow u =31.38\ yard/s\\\Rightarrow u =31.38\times 3\ ft/s\\\Rightarrow u =94.13\ ft/s

Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.

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