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2 years ago

15

A car traveling initially at 9.49 m/s accelerates at the rate of 0.988 m/s^2 for 3.05s. What is it’s velocity at the end of the

acceleration?

2 answers:

krok68 [10]2 years ago

7 0

Answer:

• From first equation of motion:

${ \boxed{ \bf{v = u + at}}}$

substitute:

$v = 9.49 + (0.988 \times 3.05) \\ v = 9.49 + 3.0134 \\ v = 12.50 \: \: m {s}^{ - 1}$

weeeeeb [17]2 years ago

6 0

Answer:

12.50 m/s

Explanation:

Vi = 9.49 m/s

a = 0.988 m/s²

t = 3.05 s

Vf = ?

Vf = Vi + at

Vf = 9.49 + (0.988)(3.05)

Vf = 12.50 m/s

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Answer:

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Explanation:

Given that,

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Using formula of torque

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Put the value into the formula

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Using formula of torque

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$\tau_{3}=(F_{3}\sin45+F_{3}\cos45)\times\dfrac{a}{2}$

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Explanation:

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Answer:

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