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Elan Coil [88]
3 years ago
11

A 3.5 kg wooden block is pulled with 12.5 N of force across a rough surface. If the force of friction is 4.8 N, find the net for

ce and the acceleration of the block. (Start by drawing a force diagram.)
Physics
1 answer:
Vesnalui [34]3 years ago
3 0

Answer:

F = 7.7 [N]

a = 2.2 [m/s²]

Explanation:

In the attached picture we see the forces on the wooden block, the total force is obtained by means of the sum of the pushforce and the friction force.

F_{total}=12.5-4.8\\F_{total}=7.7[N]

Now we must apply Newton's second law, which tells us that the total force on the body must be equal to the product of mass by acceleration.

F=m*a

7.7=3.5*a\\a=2.2[m/s^{2} ]

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a) F = 660.576\,N, b) a_{c} = 6.953\,\frac{m}{s^{2}}, c) v \approx 7255.423\,\frac{m}{s}, \omega = 9.583\times 10^{-4}\,\frac{rad}{s}, d) T \approx 1.821\,h

Explanation:

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F = G\cdot \frac{m\cdot M}{r^{2}}

F = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot \frac{(95\,kg)\cdot (5.972\times 10^{24}\,kg)}{(7.571\times 10^{6}\,m)^{2}}

F = 660.576\,N

b) The centripetal acceleration of the satellite is:

a_{c} = \frac{660.576\,N}{95\,kg}

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c) The speed of the satellite is:

v = \sqrt{a_{c}\cdot R}

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\omega = \frac{7255.423\,\frac{m}{s} }{7.571\times 10^{6}\,m}

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d) The period of the satellite's rotation around the Earth is:

T = \frac{2\pi}{\left(9.583\times 10^{-4}\,\frac{rad}{s} \right)} \cdot \left(\frac{1\,hour}{3600\,s} \right)

T \approx 1.821\,h

6 0
3 years ago
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