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Wittaler [7]
3 years ago
9

a body weights 28N at a height of 3200km from the earth surface.What will be the gravitational force on that body if its lies on

the earth surface.
Physics
1 answer:
alekssr [168]3 years ago
4 0

Answer:

The object would weight 63 N on the Earth surface

Explanation:

We can use the general expression for the gravitational force between two objects to solve this problem, considering that in both cases, the mass of the Earth is the same. Notice as well that we know the gravitational force (weight) of the object at 3200 km from the Earth surface, which is (3200 + 6400 = 9600 km) from the center of the Earth:

F_G=G\,\frac{M_E\,m}{d^2} \\28\,\,N=G\,\frac{M_E\,m}{9600000^2}

Now, if the body is on the surface of the Earth, its weight (w) would be:

F_G=G\,\frac{M_E\,m}{d^2} \\w=G\,\frac{M_E\,m}{6400000^2}

Now we can divide term by term the two equations above, to cancel out common factors and end up with a simple proportion:

\frac{w}{28} =\frac{9600000^2}{6400000^2} \\\frac{w}{28} =\frac{9}{4} \\\\ \\w=\frac{9\,*\,28}{4}\,\,\,N\\w=63\,\,N \\

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Answer:

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2 years ago
If you want to double the kinetic energy of a gas molecule, by what factor must you increase its momentum?A) 16B) 2^(1/2)
worty [1.4K]

Answer: \sqrt{2}

The linear momentum p is given by the following equation:

p=m.v   (1)

Where m is the mass and v the velocity.

On the other hand, the kinetic energy K is given by:

K=\frac{p^{2}}{2m}   (2)

Which is the same as:

K=\frac{1}{2}m.v^{2}

Now, if we double the kinetic energy, equation (2) changes to:

2K=2\frac{p^{2}}{2m}  

2K=\frac{p^{2}}{m}   (3)

So, if we want to obtain the kinetic energy as shown in (3), the only option that works is increasing momentum by a factor of \sqrt{2} or 2^{1/2}:

Applying this in (2):

K=\frac{(\sqrt{2}p)^{2}}{2m}

K=\frac{(2p)^{2}}{2m}

K=\frac{p^{2}}{m}>>>As we can see, this equation is the same as equation (3)

Therefore, the correct answer is B

3 0
3 years ago
Why can acceleration include only a change in direction?
Fynjy0 [20]

First we shall look at some terms:

Velocity - frequency of change of an objects position.

Acceleration - frequency of change of an objects velocity


Since velocity covers both change in direction and speed, we can asume that the answer is B, not because it's the only one we worked with, but because the others don't make any sense:

A - this is not correct, because a chnage in direction can be done withough a change in speed.

C - The scientific definition of acceleration does include speed, and this is not a reasonable answer.

D - Velocity can change, unlike said in the sentence.


Hope it helped,


BioTeacher101

6 0
4 years ago
Read 2 more answers
What is the relationship between the weight of an object and the radius of the earth​
joja [24]

Explanation:

if radius is greater than the force acting on that object is more and if the radius is smaller then the force acting on it is less.

7 0
3 years ago
A hoop, a solid cylinder, a spherical shell, and a solid sphere are placed at rest at the top of an inclined plane. All the obje
iogann1982 [59]

Answer:

Solid sphere will reach first

Explanation:

When an object is released from the top of inclined plane

Then in that case we can use energy conservation to find the final speed at the bottom of the inclined plane

initial gravitational potential energy = final total kinetic energy

mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

now we have

I = mk^2

here k = radius of gyration of object

also for pure rolling we have

v = R\omega

so now we will have

mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mk^2)(\frac{v^2}{R^2})

mgh = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})

v^2 = \frac{2gh}{1 + \frac{k^2}{R^2}}

so we will say that more the value of radius of gyration then less velocity of the object at the bottom

So it has less acceleration while moving on inclined plane for object which has more value of k

So it will take more time for the object to reach the bottom which will have more radius of gyration

Now we know that for hoop

mk^2 = mR^2

k = R

For spherical shell

mk^2 = \frac{2}{3}mR^2

k = \sqrt{\frac{2}{3}} R

For solid sphere

mk^2 = \frac{2}{5}mR^2

k = \sqrt{\frac{2}{5}} R

So maximum value of radius of gyration is for hoop and minimum value is for solid sphere

so solid sphere will reach the bottom at first

7 0
3 years ago
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