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MA_775_DIABLO [31]
3 years ago
14

What is the ratio of the gravitational force on an astronaut when he is in a space satellite in a circular orbit 315 km above th

e earth to the gravitational force on the astronaut when he is on the surface of the earth? The earth’s diameter is 12760 km.
Physics
1 answer:
Nina [5.8K]3 years ago
8 0

Answer:

The ratio of gravitational force is 0.8920

Explanation:

The gravitational force is given by  F=\frac{GMm}{R^{2} }

  The gravitational force on the astronaut when he is in

           Surface of earth  F₁ =G\frac{Mm}{R^{2} }        eqn 1

          Circular orbit F₂ =G\frac{Mm}{(R+h)^{2} }           eqn 2

Where M is mass of earth,

           m is mass of astronaut

          R is radius of earth 6380 km,

         h is the distance from earth surface 315 km.

Dividing eqn 1 and 2

                   =    \frac{R^{2} }{(R+h)^{2} }

                   =\frac{(6380)^{2} }{(6380+315)^{2} }  

                     =0.8920

                  The ratio of gravitational force is 0.8920

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If we take a look at the estimates of <span>World Wide Fund for Nature, an organization that works toward combating species extinction, their estimates vary from 200 to 100 000 - but a probable number is 20 000 (d). </span>
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A box is at rest on a table. What can you say about the forces acting on the box?
Nikitich [7]
You can tell a lot about an object that's not moving,
and also a lot about the forces acting on it:

==> If the box is at rest on the table, then it is not accelerating.

==> Since it is not accelerating, I can say that the forces on it are balanced.

==> That means that the sum of all forces acting on the box is zero,
and the effect of all the forces acting on it is the same as if there were
no forces acting on it at all.

==> This in turn means that all of the horizontal forces are balanced,
AND all of the vertical forces are balanced.

Horizontal forces:
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All of the forces on this list must add up to zero. So ...

(sliding friction force) = (pushing force), in the opposite direction.

If nobody pushing the box, then sliding friction force = zero.

Vertical forces:
gravitational force (weight of the box, pulling it down)
normal force (table pushing the box up)

All of the forces on this list must add up to zero, so ...

(Gravitational force down) + (normal force up) = zero

(Gravitational force down) = -(normal force up) .
6 0
3 years ago
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Please help calculate them<br> A-f please very urganr
valentina_108 [34]
Three 40w lamp for 6 hours
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3 years ago
Sunlight reflects from a concave piece of broken glass, converging to a point 34 cm from the glass. what is the radius of curvat
wolverine [178]
The rays of light coming from the Sun are parallel to each other, so when they are reflected by the concave piece of glass (which acts as a concave mirror) they converge into the focus of the mirror, which is
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6 0
3 years ago
What is the moment of inertia of a 2.0 kg, 20-cm-diameter disk for rotation about an axis (a) through the center, and (b) throug
FinnZ [79.3K]

Answer:

(a) I=0.01 kg.m²

(b) I=0.03 kg.m²

Explanation:

Given data

Mass of disk M=2.0 kg

Diameter of disk d=20 cm=0.20 m

To Find

(a) Moment of inertia through the center of disk

(b) Moment of inertia through the edge of disk

Solution

For (a) Moment of inertia through the center of disk

Using the equation of moment  of Inertia

I=\frac{1}{2}MR^{2}\\  I=\frac{1}{2}(2.0kg)(0.20m/2)^{2}\\  I=0.01 kg m^{2}

For (b) Moment of inertia through the edge of disk

We can apply parallel axis theorem for calculating moment of inertia

I=(1/2)MR^{2}+MD\\ Here\\D=R\\I=(1/2)(2.0kg)(0.20m/2)^{2}+(2.0kg)(0.20m/2)^{2}\\  I=0.03kgm^{2}

8 0
3 years ago
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