A certain force exerted for 1.2s raises the speed of an object from 1.8m/s to 4.2m/s . Find its acceleration if mass of the object is5.5kg . Calculate the force applied.
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Answered by
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Time = 1.2 second
Initial velocity = 1.8m/sec
Final velocity = 4.2m/sec
Now,
Acceleration = final v - initial v/time
So,
![a = \frac{4.2 - 1.8}{1.2} = \frac{2.4}{1.2} = 2 a = \frac{4.2 - 1.8}{1.2} = \frac{2.4}{1.2} = 2](https://tex.z-dn.net/?f=a+%3D++%5Cfrac%7B4.2+-+1.8%7D%7B1.2%7D++%3D++%5Cfrac%7B2.4%7D%7B1.2%7D++%3D+2+)
So,
Acceleration = 2m/sec²
Mass (given) = 5.5 kg
Now,
![force = mass \times acceleration force = mass \times acceleration](https://tex.z-dn.net/?f=force+%3D+mass+%5Ctimes+acceleration)
So,
F= 5.5 * 2 = 11N
Hope this will be helping you
WARM REGARDS
Sahil khirwal
Initial velocity = 1.8m/sec
Final velocity = 4.2m/sec
Now,
Acceleration = final v - initial v/time
So,
So,
Acceleration = 2m/sec²
Mass (given) = 5.5 kg
Now,
So,
F= 5.5 * 2 = 11N
Hope this will be helping you
WARM REGARDS
Sahil khirwal
Answered by
0
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