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3 years ago

1) What is the weight of a 356kg object?

Physics

1 answer:

stiks02 [169]3 years ago

4 0

1) 3489 N

2) The microbumps over the surfaces

3) See explanation

Explanation:

The gravitational force acting on an object (also known as weight of the object) is the attractive force with which the Earth pulls the object towards the ground.

It can be calculated using the formula:

$W=mg$

where

m is the mass of the object

g is the acceleration due to gravity

In this problem, we have:

m = 356 kg is the mass of the object

$g=9.8 m/s^2$ is the acceleration due to gravity

Therefore, the weight of the object is:

$W=(356)(9.8)=3489 N$

The force of friction is a force that acts whenever there is an object sliding over a surface. It is a resistive force, therefore its direction is always opposite to the direction of motion of the object.

On a microscopic scale, the reason of this force of friction is the presence of "microbumps" over the surface of the sliding object (and over the surface of the floor). Because of the roughness of the two surfaces, the molecules of the two surfaces "interact" and cause a resistive force that opposes the relative motion of the object over the surface.

Macroscopically, for an object over a flat surface, the magnitude of the force of friction is

$F_f=\mu mg$

where

$\mu$ is the coefficient of friction

m is the mass of the object

g is the acceleration due to gravity

There are two types of force of friction:

- Force of static friction: this occurs when the object is still at rest. We experience this force when we try to put in motion an object on the ground, by applying a push: we notice that if the force we apply is not strong enough, the object will remain at rest. This is because there is a static force of friction acting on the object, that equals the force of push that we are applying, but in the opposite direction. The magnitude of the force of friction can vary up to a maximum value, given by

$F_{max}=\mu_s mg$

where

$\mu_s$ is the coefficient of static friction

m is the mass of the object

g is the acceleration due to gravity

- Force of kinetic friction: this is the force of friction that occurs when the object is already in motion. The kinetic force of friction opposes the motion of the object, and its magnitude is

$F_k = \mu_k mg$

where

$\mu_k$ is the coefficient of kinetic friction

Generally, for most surfaces, $\mu_k < \mu_s$ , therefore the force of kinetic friction is generally less than the force of static friction.

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A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

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Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

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Where:

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$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

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Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

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If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

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$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

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