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3 years ago

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An object acted on by three forces moves with constant velocity. One force acting on the object is in the positive xx direction

and has a magnitude of 6.9 N; a second force has a magnitude of 4.5 N and points in the negative y direction. Find the direction and magnitude of the third force acting on the object.

1 answer:

Rudik [331]3 years ago

8 0

Answer with Explanation:

We are given that

$F_1=6.9 N$

$F_2=4.5 N$

We have to find the direction and magnitude of the third force acting on the object.

Resultant force,F= $\sqrt{F^2_1+F^2_2}$

$F=\sqrt{(6.9)^2+(4.5)^2}=8.24 N$

The object moves with constant velocity .Therefore, net force on object is equal to zero

So,Third force, $F_3=F=8.24 N$

Direction, $\theta=tan^{-1}(\frac{F_2}{F_1})$

$\theta=tan^{-1}(\frac{4.5}{6.9})=33.02^{\circ}$

Angle lies in second quadrant because the direction of third force is opposite to the direction of the resultant force of F1 and F2.

Therefore, $\theta=\pi-\theta=180-33.02=146.98^{\circ}$

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Answer

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Answer:

<h3>The answer is 8 kg</h3>

Explanation:

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$m = \frac{40}{5} \\$

We have the final answer as

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3 years ago

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Answer:

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Explanation:

Solution:

Data Given:

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A = width x Length

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U = 1/2 C Δ $v^{2}$

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Solving for L, we get:

L = 1.11 x $10^{6}$ m,

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