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2 years ago

By what factor would your weight be multiplied if the earth were1/2 as massavise and the diameter was unchanged

1 answer:

Nutka1998 [239]2 years ago

4 0

<span>Let F be the force of gravity, G be the gravitational constant, M be the mass of the earth, m your mass and r the radius of the earth, then:

F = G(Mm / (4(pi)*r^2))

The above expression gives the force that you feel on the earth's surface, as it is today!

Let us now double the mass of the earth and decrease its diameter to half its original size.

This is the same as replacing M with 2M and r with r/2.

Now the gravitational force (F' ) on the new earth's surface is given by:

F' = G(2Mm / (4(pi)(r/2)^2)) = 2G(Mm / ((1/4)*4(pi)*r^2)) = 8G(Mm / (4(pi)*r^2)) = 8F

So:

F' = 8F

This implies that the force that you would feel pulling you down (your weight) would increase by 800%!

You would be 8 times heavier on this "new" earth!</span>

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A 3.2-kg thin, circular hoop with a radius of 5.4 m is rotating about an axis through its center and perpendicular to its plane.

elena-14-01-66 [18.8K]

Answer:

Torque = –207.4 Nm

Explanation:

Given M = 3.2kg, r = 5.4m, α = –12rad/s² (it is slowing down)

Torque = I × α

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I = moment of inertia

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Torque = 3.2×5.4×(– 12)

Torque = –207.4 Nm

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3 years ago

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A constant net force of 20 N is applied to a 32kg car, causing it to speed up from 4.0 to 9.0 m/s. How long is the force applied

pishuonlain [190]

Answer:

8 seconds

Explanation:

From Newton's second law;

Ft = m(v-u)

F = Force applied

t = time taken

v = final velocity

u = initial velocity

20 * t = 32 (9 - 4)

20t = 32 * 5

t = 32 * 5/ 20

t = 8 seconds

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2 years ago

Suppose that in a lightning flash the potential difference between a cloud and the ground is 1.0*109 V and the quantity of charg

nata0808 [166]

Answer:

a) $U_{e} = 3 \times 10^{10}\,J$ , b) $v \approx 7745.967\,\frac{m}{s}$

Explanation:

a) The potential energy is:

$U_{e} = Q \cdot \Delta V$

$U_{e} = (30\,C)\cdot (1.0\times 10^{9}\,V)$

$U_{e} = 3 \times 10^{10}\,J$

b) Maximum final speed:

$U_{e} = \frac{1}{2}\cdot m \cdot v^{2}\\v = \sqrt{\frac{2\cdot U_{e}}{m} }$

The final speed is:

$v=\sqrt{\frac{2\cdot (3 \times 10^{10}\,J)}{1000\,kg} }$

$v \approx 7745.967\,\frac{m}{s}$

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3 years ago

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The core of a star must be at the temperature of 10,000,000 degrees Celsius for hydrogen fusion to begin.

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two charges are placed at corners A and B of a square of side length. How much work is needed to move a charge from point C to D

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