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4 years ago

What would happen to the moon if earth stopped exerting the force of gravity on it?

2 answers:

6 0

There are two equal forces of gravity between the Earth and the Moon.
One force pulls the Moon toward the Earth.
The other force pulls the Earth toward the Moon.

If only this gravity suddenly switched off, then the moon would
continue to orbit the Sun, very much as it does now.

If ALL gravity suddenly switched off, then . . .

-- the Moon would stop orbiting the Earth and would sail away, in
a straight line and at the speed it had when gravity disappeared;

-- the Earth would stop orbiting the Sun and would sail away, in
a straight line and at the speed it had when gravity disappeared;

-- all the gases surrounding the Earth ... which we call "air" ... would
start drifting away, and expanding into a giant cloud of gas, and stop
being an atmosphere;

-- the Sun would completely fall apart, expand into a giant cloud of gas,
and stop being a star.

larisa86 [58]4 years ago

3 0

The moon will stop orbiting and will fly away and never stop running

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A magnetic field of 0.90 T is aligned vertically with a flat plane. The magnetic flux through the plane is 7.3 × 10^-3 Wb. What

alexandr402 [8]

Answer:

0.10 m

Explanation:

The magnetic flux through the plane is given by

$\Phi = BA$

where

B is the magnetic field intensity

A is the area of enclosed by the pipe

In this problem, we know

$\Phi = 7.3\cdot 10^{-3} Wb$ is the flux

B = 0.90 T is the magnetic field strength

Solving the equation for A, we find the area enclosed by the pipe

$A=\frac{\Phi}{B}=\frac{7.3\cdot 10^{-3} Wb}{0.90 T}=8.1\cdot 10^{-3} m^2$

We know that the area is given by

$A=\pi r^2$

where r is the radius. Solving for r, we find the radius:

$r=\sqrt{\frac{A}{\pi}}=\sqrt{\frac{8.1\cdot 10^{-3} m^2}{\pi}}=0.05 m$

And so the diameter is twice the radius:

$d=2r=2(0.05 m)=0.10 m$

5 0

3 years ago

A tall cylinder contains 30 cm of water. Oil is carefully poured into the cylinder, where it floats on top of the water, until t

Anastaziya [24]

Answer:

The gauge pressure is $P_g = 2058 \ P_a$

Explanation:

From the question we are told that

The height of the water contained is $h_w = 30 \ cm = 0.3 \ m$

The height of liquid in the cylinder is $h_t = 40 \ cm = 0.4 \ m$

At the bottom of the cylinder the gauge pressure is mathematically represented as

$P_g = P_w + P_o$

Where $P_w$ is the pressure of water which is mathematically represented as

$P_w = \rho_w * g * h_w$

Now $\rho_w$ is the density of water with a constant values of $\rho_w = 1000 \ kg /m^3$

substituting values

$P_w = 1000 * 9.8 * 0.3$

$P_w = 2940 \ Pa$

While $P_o$ is the pressure of oil which is mathematically represented as

$P_o = \rho_o * g * (h_t -h_w )$

Where $\rho _o$ is the density of oil with a constant value

$\rho _o = 900 \ kg / m^3$

substituting values

$P_o = 900 * 9.8 * (0.4 - 0.3)$

$P_o = 882 \ Pa$

Therefore

$P_g = 2940 - 882$

$P_g = 2058 \ P_a$

6 0

3 years ago

The amount of friction divided by the weight of an object forms a unit less number called the

Romashka [77]

Answer:

Coefficient of friction.

Explanation:

The amount of friction divided by the weight of an object is equal to the coefficient of friction. It is a dimensional less number. It can be given by :

$F=\mu N$