Physics, asked by TrapNation, 1 year ago

A proton in a cyclotron changes its velocity from 30 km/s north to 40km/s east in 20s. What is the magnitude of average acceleration during this time ???

Answers

Answered by Anonymous
397
Bonjour!

The magnitude of average acceleration during this time will be 2.5 km/s².

See the attached picture for calculations.

Hope this helps...:)
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Answered by lidaralbany
31

Answer: The average acceleration is a = 2.5 km/s^{2}

Explanation:

Given that,

Velocity  v_{1} = -30 km/s from north

Velocity  v_{2} = 40 km/s to east

Time t = 20 s

The resultant velocity is

v = \sqrt{v_{1}^{2}+v_{2}^{2}}

v = \sqrt{40^{2}+30^{2}}

v = 50 km/s

Now, the average acceleration is

Acceleration = resultant velocity/ time

a = \dfrac{v}{t}

a = \dfrac{50 km/s}{20s}

a = 2.5 km/s^{2}

Hence, the average acceleration is a = 2.5 km/s^{2}.

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