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3 years ago

What is the average speed, over the first 4.0 s of its motion, of a pebble released from rest off a bridge?

Physics

1 answer:

galben [10]3 years ago

8 0

Answer:

Average speed, over the first 4.0 s of its motion, of a pebble released from rest off a bridge = 19.6 m/s

Explanation:

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case the initial velocity of body = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate displacement when t = 4 seconds.

Substituting

$S=0*t+\frac{1}{2} *9.8*4^2\\ \\ S=78.4 meter$

So a distance of 78.4 meter has been traveled by pebble in 4 seconds.

Average speed = 78.4/4 = 19.6 m/s

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snow_tiger [21]

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3 years ago

A force, F1, of magnitude 2.0 N and directed due east is exerted on an object. A second force exerted on the object is F2 = 2.0

IgorLugansk [536]

Answer:

Magnitude of the force is

$F_3 = 2.83$

direction of the force is given as

$\theta = 45 degree$ West of South

Explanation:

As we know that force is a vector quantity and in order to find the resultant of two or more forces we need to add them vectorialy

So here we have

$\vec F_1 + \vec F_2 + \vec F_3 = 0$

here we know that first force is of magnitude 2 N towards east

$\vec F_1 = 2 \hat i N$

second force is also of 2.0 N due North

$\vec F_2 = 2 \hat j$

now from above equation

$2\hat i + 2\hat j + \vec F_3 = 0$

$\vec F_3 = -2\hat i - 2\hat j$

so magnitude of the force is given as

$F_3 = \sqrt{2^2 + 2^2}$

$F_3 = 2.83$

direction of the force is given as

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$\theta = tan^{-1}\frac{-2}{-2}$

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3 years ago

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Explanation:

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3 years ago

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Answer:

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