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PolarNik [594]
3 years ago
8

Question 1

Physics
1 answer:
valentina_108 [34]3 years ago
6 0

The distance covered is simply the length of the entire trip, which is 12m + 16m, or 28m.

The displacement is the distance from the starting point to the ending point along with the direction of the net motion. The dog walks 12m east then 16m west, so its resultant displacement is 4m west.

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A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s (~ 18 mph). What is the total time it is i
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Answer:

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To find:

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t = \frac{v-u}{a}

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Acceleration of gravity is taken as negative because ball is moving in opposite direction.

Solution:

A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s.

The time taken by the ball to reach the maximum height is given by,

t = \frac{v-u}{a}

Where t = time to reach maximum height

v = final velocity of the ball = 0 m/s

u = initial velocity of ball = 8 m/s

a = acceleration due to gravity = -9.8

Acceleration of gravity is taken as negative because ball is moving in opposite direction.

t = \frac{0-8}{-9.8}

t = 0.8163 s

Thus, time taken by the ball to reach the ground again = time taken to reach maximum height

So, Total time required for ball to reach ground = 2t = 2 × 0.8163

Total time required for ball to reach ground = 1.6326 s

The total time it is in the air for the ball is 1.6326 s

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