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belka [17]
3 years ago
5

A constant force of 12N is applied for 3.0s to a body initially at rest. The final velocity of the body is 6.0ms–1. What is the

mass of the body?
Physics
1 answer:
sp2606 [1]3 years ago
4 0
From the question,
u = 0m {s}^{ - 1}
v = 6m {s}^{ - 1}

t = 3s
F=12N



Using Impulse, the product of the constant force, F and time t equals the product of the mass of the body and change in velocity.

Ft =m(v-u)


12(3.0)=m(6.0- \: 0)
This implies that

36.0 = 6m
m =  \frac{36.0}{6.0}
\therefore \: m = 6.0kg


You can also use the equation of linear motion,
v = u + at
6 = 0 + a(3)
6 = 3a
a =  \frac{6}{3}

a = 2 {ms}^{ - 2}
But
F=ma
12 = m(2)
12 = 2m
\frac{12}{2}  = m
\therefore \: m = 6kg
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Answer:

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Explanation:

Using kinematics equations:

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Use \Delta y = 0 due to condition of distance traveled.

Solving second equation for time, there are two solutions. t=0 and

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Use the expression in the first equation to have

R = \frac{2v^2 \cos\theta\sin\theta}{g}

Using trigonometric identities, you have the answer of the distance.

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A 1000-kg whale swims horizontally to the right at a speed of 6.0 m/s. It suddenly collides directly with a stationary seal of m
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Answer:

1.27\times 10^{12}\Omega/m

Explanation:

We are given that

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Thickness=1\mu m

Radius=r=\frac{d}{2}=\frac{1}{2}\mu m=0.5\times 10^{-6} m

Using 1\mu m=10^{-6} m

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Using \pi=3.14

Internal resistance per unit length=1.27\times 10^{12}\Omega/m

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