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MakcuM [25]
3 years ago
5

A constant force of 4.5N acts on a 7.2 kg object for 10 s what is the change in the object velocity

Physics
1 answer:
nata0808 [166]3 years ago
7 0

Answer:

<em>The change in the object's velocity is 6.25 m/s</em>

Explanation:

<u>Force </u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = ma      [1]

Where a is the acceleration of the object.

On the other hand, the equations of the Kinematics describe the motion of the object by the equation:

v_f=v_o+at\qquad\qquad[2]

Where:

vf is the final speed

vo is the initial speed

a is the acceleration

t is the time

It's given a constant force of F=4.5 N acts on an object of mass m=7.2 kg. The object has an acceleration we can calculate solving [1] for a:

a=\frac{F}{m}

a=\frac{4.5}{7.2}=0.625

a=0.625\ m/s^2

It's required con compute the change of velocity (or its magnitude, the speed). From [2]:

v_f-v_o=at

The expression at the left side is the change of speed Δv:

\Delta v=at

Knowing the time is t=10 s:

\Delta v=0.625*10

\Delta v=6.25\ m/s

The change in the object's velocity is 6.25 m/s

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