Answered

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A 3.0 kg rock is dropped from rest. What will its velocity be in 6.0 s if a 7.0 N force of air resistance acts on it?

Sagot :

miroslavass

Answer:

14 m/s^2

Explanation:

[tex]v_{f}=V_{i} +at[/tex]

[tex]V_{f}=[/tex]final velocity

[tex]V_{i}=[/tex]initial velocity

a= acceleration

t=time

Initial velocity is zero, since the rock was originally at rest and time is 6 seconds.

You are looking for the final velocity, but you need acceleration to find it. Remember that:

[tex]F=m*a[/tex]

Where F is force, m is mass, and a is acceleration. You know force is 7.0 N and mass is 3.0 kg. Use these to find a.

[tex]7.0 N= 3*a[/tex]

[tex]\frac{7}{3} = a[/tex]

So now...

[tex]v_{f}=V_{i} +at[/tex]

[tex]v_{f}=0 +(\frac{7}{3}) (6)[/tex]

[tex]v_{f}= 14 m/s^{2}[/tex]

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