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11111nata11111 [884]

3 years ago

7

A block is moving at constant speed due to a horizontal force pulling to the right. The coefficient of kinetic friction, Hk, bet

ween the block and the surface is 0.20 and the mag- nitude of the frictional force is 100.0 N, what is the weight of the block? (a 400 N (b) 600 N (c) 500 N (d) 267 N

1 answer:

VashaNatasha [74]3 years ago

4 0

Answer:

c) 500 N

Explanation:

If the block moves with constant speed there is no acceleration. We draw a free body diagram and define the forces on the body.

$F_{f}=FrictionForce\\F_{e}=ExternalForce\\N=NormalForce\\W=Weight\\$

The equation for the frictional force is:

$F_{f}=N*H_{k}$

Where $H_{k}$ is the kinetic friction coefficient. We write the equilibrium equation in the y direcction:

$\sum F_{y}\rightarrow N-W=0\\N=W$

We replace this result in the equation for the frictional force:

$F_{f}=W*H_{k}$

We replace the data given by the exercise and find the weight

$W=\frac{F_{f}}{H_{k}}\\W=\frac{100}{0.2}=500\: N$

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The critical velocity for a satellite orbiting around a planet is given by

$v=\sqrt{\frac{GM}{R+h}}$

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$M=5.98\cdot 10^{24} kg$ (mass of the Earth)

$R=6.37\cdot 10^6 m$ (Earth's radius)

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Substituting,

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The period of the orbit is given by the circumference of the orbit divided by the velocity:

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$h=2.60\cdot 10^6 m$

v = 6668 m/s

Substituting,

$T=\frac{2\pi (6.37\cdot 10^6 + 2.60\cdot 10^6)}{6668}=8452 s$

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One day consists of:

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Learn more about gravitational force:

brainly.com/question/1724648

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