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Margarita [4]
3 years ago
5

Suppose we could shrink the Earth without changing its mass. At what fraction of its current radius would the free-fall accelera

tion at the surface be three times its present value?
Physics
2 answers:
spin [16.1K]3 years ago
7 0

Answer:

R' = \frac{1}{\sqrt{3}}R

Explanation:

The acceleration due to gravity on the surface of the Earth is given by:

g=\frac{GM}{R^2}

where

G is the gravitational constant

M is the mass of the Earth

R is the radius of the Earth

Here we want to find the new Earth radius R' for which the gravitational acceleration at the surface, g', would be 3 times the current value of g:

g' = 3g

So we would have

\frac{GM}{R'^2}=3(\frac{GM}{R^2})

Solving the equation for R', we find

R'^2 = \frac{1}{3}R^2\\R' = \frac{1}{\sqrt{3}}R

musickatia [10]3 years ago
5 0

The earth can lose and gain weight.

<h2>Further Explanation </h2>

This may be a subject that is rarely discussed or even known by people. The interesting fact is that the Earth can actually lose mass and increase mass caused by several things. Basically every day is very possible for the Earth to collide with dust, comets, and meteors that can penetrate the atmosphere and land on Earth. This, of course, can indirectly increase its mass. On the other hand, mass loss occurs when there is a gas released from the atmosphere out so that the mass decreases when it occurs. Dr.a Chris Smith once tried to calculate this and it is estimated that each year 40,000 tons of space dust increases the mass of the Earth, but 95,000 tons of hydrogen and 1,600 tons of helium are released into space. On the other hand, it takes trillions of years for the hydrogen on this planet to run out but in contrast to helium which cannot be seen and is calculated in exact amounts. Some people have even warned that the Earth could run out of helium in the next three decades with the current conditions we are wasting on big projects.

If the size of the earth decreases, the force of gravity will decrease, the earth rotates faster, with lighter gravity, people will become lighter, faster, and have higher posture due to the lack of gravitational pressure.

Learn more

what if we shrink the earth  brainly.com/question/12644915

Details

Grade:  High School

Subject:  Physics

keywords: shrink, earth.

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A 110 kg ice hockey player skates at 3.0 m/s toward a railing at the edge of the ice and then stops himself by grasping the rail
velikii [3]

Answer:

Δ KE = -495 J

Explanation:

given,

mass of the ice hockey player = 110 Kg

initial speed = 3 m/s

final speed  = 0 m/s

distance, d = 0.3 m

change in kinetic energy

\Delta K E = \dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2

\Delta K E = \dfrac{1}{2}mv(0)^2 - \dfrac{1}{2}\times 110\times 3^2

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7 0
3 years ago
A maintenance worker wants to torque an engine bolt to 65.0 N m. If the torque wrench is 35cm in length, what is force applied t
harina [27]

Answer:

Force = 186 N

Explanation:

Torque is the rotational equivalent of linear force. It can be easely calculated using the formula :

Torque = \vec{r} \times \vec{F}

Where \vec{r} is a vector that from the origin of the coordinate system to the point at which the force is applied (the position vector), \vec{F} is the applied force.

The easiest way of computing the force is by setting the origin of the coordinate system to the lowest point of the torque wrench.  By doing this we have that r (the magnitud of the position vector) is 35cm.

Before computing the force we need to set all our values to the international system of units (SI). The torque is already in SI. The one missing is the length of the torque wrench (it is in centimeters and we need it in meters). So :

35cm * \dfrac{1m}{100cm} = 0.35m

Now using the torque formula:

Torque = \vec{r} \times \vec{F}

Torque = rFsin(\theta)}

Where \theta is the smaller angle between the force and the position vector. Because the force is applied perpendiculary to the position vector  \theta = 90°, thus :

Torque = rFsin(\theta)}

65 N m = (0.35m)Fsin(90°)}

F = \dfrac{65N m} {(0.35m)sin(90°)}

F = \dfrac{65N m} {(0.35m)sin(90°)}

F = \dfrac{1300} {7}N

so the force is approximately 186 N.

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One gram of pure gold would contain lots of gold atoms- A.

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