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3 years ago

7

A car is initially moving at 10.5 m/s and accelerates uniformly to reach a speed of 21.7 m/s in 4.34 s. How far did the car move

during this acceleration?
A. 69.9m
B. 6.99m
C. 69.6m/s
D. 6.99m/s

1 answer:

Zarrin [17]3 years ago

3 0

Answer:

A. 69.9m

Explanation:

Given parameters:

Initial velocity = 10.5m/s

Final velocity = 21.7m/s

Time = 4.34s

Unknown:

Distance traveled = ?

Solution:

Let us first find the acceleration of the car;

Acceleration = $\frac{v - u}{t}$

v is final velocity

u is initial velocity

t is the time

Acceleration = $\frac{21.7 - 10.5}{4.34}$ = 2.58m/s²

Distance traveled;

V² = U² + 2aS

21.7² = 10.5² + 2 x 2.58 x S

360.64 = 2 x 2.58 x S

S = 69.9m

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From the question we are told

Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e

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2 years ago

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