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3 years ago

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The gravitational pull of Jupiter is greater than the gravitational pull of Earth. How would adding a “JUPITER MODE” to the virt

ual activity change the results of Galileo’s experiment?

1 answer:

Vlad1618 [11]3 years ago

5 0

The Galileo's experiment is about Earth’s gravitational force have on objects of different masses.

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You and your friends are having a discussion about weight. He/she claims that he/she weighs less on the 100th floor of a buildin

Viktor [21]

Answer:

if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

Explanation:

The weight of a person in the force with which the Earth attracts the person, therefore can be calculated using the law of universal attraction

F = G m M / r²

Where m is the mass of the person, M the masses of the earth

Let's call the person's weight at ground level as Wo and suppose the distance to the center of the Earth is Re

W₀ = G m M / Re²

In the calculation of the weight of the person on the 100th floor the only thing that changes is the distance

r = Re + 100 r₀

Where r₀ is the distance between the floors, which is approximately 2.5 m, so the distance is

r = Re + 250

We substitute

W = G m M / r²

W = G m M / (Re + 250)²

The value of Re is 6.37 10⁶ m, so we can take it out as a factor and perform a serial expansion of the remaining fraction

W = G m M / Re² (1+ 250 / Re)²

(1 + 250 / Re)⁻² = 1 + (-2) 250 / Re + (-2 (-2-1)) / 2 (250 / Re)² +….

The value of the expression is

(1 + 250 / Re)⁻² = 1 -2 250 / 6.37 10⁶ -30 (250 / 6.37)² 10⁻¹² + ...

We can see that the quadratic term is very small, which is why we despise it, we substitute in the weight equation

W = G m M / Re² (1 - 78.5 10⁻⁶)

Remains

W = Wo (1 - 7.85 10⁻⁵)

We can see that if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

4 0

3 years ago

Read 2 more answers

mass and weight are similar, but not the same thing. In which of the following examples would the objects weight change, but mas

OverLord2011 [107]

When an astronaut travels from the earth to the moon, her weight changes, but her mass remains constant. <em>(C ).</em>

7 0

3 years ago

You could use newton’s second law to calculate the force applied to an object if you knew the objects mass and its _____.

olga2289 [7]

Answer:

You could use newton’s second law to calculate the force applied to an object if you knew the objects mass and its <u>acceleration.</u>

Explanation:

By, Newtons second law, the force applied on an object directly varies with the acceleration caused and the mass of the object.

This is given by :

$F=m\ a$

Where $F$ represents force applied on the object , $m$ represents mass of the object and $a$ represents the acceleration.

In order to calculate force applied on object we require the mass of the object and its acceleration. The force can be calculated by finding the product of mass and acceleration of the object.

4 0

3 years ago

In which one of the following situations is zero net work done? A) A ball rolls down an inclined plane. B) A physics student str

lidiya [134]

Answer:

Option D

Explanation:

The work done can be given by the mechanical energy used to do work, i.e., Kinetic energy and potential energy provided to do the work.

In all the cases, except option D, the energy provided to do the useful work is not zero and hence work done is not zero.

In option D, the box is being pulled with constant velocity, making the acceleration zero and thus Kinetic energy of the system is zero. Hence work done in this case is zero.

6 0

3 years ago

Read 2 more answers

A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)

kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

$W=Fd cos \theta$

where

F is the magnitude of the force applied

d is the displacement of the crate

$\theta$ is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied horizontally. Therefore, the work done is

$W=(230)(4.0)(cos 0^{\circ})=920 J$

b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is $\theta=0^{\circ}$ , since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

$W=(1300)(4.0)(cos 0^{\circ})=5200 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0

3 years ago