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7nadin3 [17]
3 years ago
9

A dragster going 15m/s increases its velocity to 25 m/s north in 4 seconds.

Physics
1 answer:
Tomtit [17]3 years ago
3 0
Initial speed, v₁ = 15 m/s
Final speed, v₂  = 25 m/s
Duration of the acceleration, t = 4 s

Calculate the acceleration.
a = (v₂ - v₁)/t
   = (25-15 m/s)/(4 s)
   = 2.5 m/s²

Answer: 2.5 m/s²
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The answer to your question is A
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Answer:

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Explanation:

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The magnitude of the acceleration of the proton is 5.74\times10^{13}\ m/s^2

(b). We need to calculate the initial peed

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v^2-u^2=2as

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0-u^2=2\times(-5.74\times10^{13})\times7.00\times10^{-2}

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u=2.83\times10^{6}\ m/s

The initial peed of the protion is 2.83\times10^{6}\ m/s

(c). We need to calculate the time interval over which the proton comes to rest

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t=\dfrac{u}{a}

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a = acceleration

Put the value into the formula

t=\dfrac{2.83\times10^{6}}{5.74\times10^{13}}

t=0.493\times10^{-7}\ sec

Hence, (a). The magnitude of the acceleration of the proton is 5.74\times10^{13}\ m/s^2

(b). The initial peed of the protion is 2.83\times10^{6}\ m/s

(c). The time is 0.493\times10^{-7}\ sec.

5 0
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Explanation:

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