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3 years ago

Which objects will likely have the greatest gravitational force between them?

Physics

1 answer:

Mariulka [41]3 years ago

3 0

Answer:

d. Two soccer balls that are touching each other

Explanation:

Let $m_1$ be the mass of a tennis ball, $m_2$ is the mass of a soccer ball.

As the mass of a soccer ball is more than the mass of a tennis ball, so

$m_2 > m_1$

Let $d_1$ be the distance between the centers of both the balls near each other and $d_2$ be the distance between the centers of both the balls touching each other.

So, $d_2 > d_1$

The gravitational force, F, between the two objects having masses M and m and separated by distance d is

$F=\frac{GMm}{d^2}$

Where G is the universal gravitational constant.

As, the gravitational force is directly proportional to the product of both the masses and inversely proportional to the square of the distance between them, so selecting the larger mass ( $m_2$ , soccer ball) separated by a lesser distance ( $d_2$ , touching) to get more gravitational force.

Therefore, there will be a larger gravitational force between them when two soccer balls touching each other.

Hence, option (d) is correct.

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Is there an eclipse happening soon in England

lyudmila [28]

A solar eclipse will be visible over a wide area of the north polar region
on Friday, March 20.

England is not in the path of totality, but it's close enough so that a large
part of the sun will be covered, and it will be a spectacular sight.

For Londoners, the eclipse begins Friday morning at 8:25 AM,when the
moon just begins to eat away at the sun's edge. It advances slowly, as more
and more of the sun disappears, and reaches maximum at 9:31 AM. Then
the obscured part of the sun begins to shrink, and the complete disk is
restored by the end of the eclipse at 10:41AM, after a period of 2 hours
16 minutes during which part of the sun appears to be missing.

The catch in observing the eclipse is:

YOU MUST NOT LOOK AT THE SUN.

Staring at the sun for a period of time can cause permanent damage to
your vision, even though you don't feel it while it's happening.

This is not a useful place to try and give you complete instructions or
suggestions for observing the sun over a period of hours. Please look
in your local newspaper, or search online for phrases like "safe eclipse
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3 0

3 years ago

Read 2 more answers

Describe how the ocean influences climate by storing heat and water? Give one example to justify your answer.

just olya [345]

The sun light that received by the water in the ocean will increase the average temperature of the water and make it warmer.
With the help of wind and current, the warm water will spread out to another region, and increasing the average temperature in that region and affecting its overall climate.
Example The temperature rise in Valdivia, Chile and in Beijing, China after receiving warm water from arctic.

7 0

3 years ago

A 0.140 kg baseball is thrown horizontally with a velocity of 28.9 m/s. It is struck with a constant horizontal force that lasts

Viefleur [7K]

Answer:

4987N

Explanation:

Step 1:

Data obtained from the question include:

Mass (m) = 0.140 kg

Initial velocity (U) = 28.9 m/s

Time (t) = 1.85 ms = 1.85x10^-3s

Final velocity (V) = 37.0 m/s

Force (F) =?

Step 2:

Determination of the magnitude of the horizontal force applied. This can be obtained by applying the formula:

F = m(V + U) /t

F = 0.140(37+ 28.9) /1.85x10^-3

F = 9.226/1.85x10^-3

F = 4987N

Therefore, the magnitude of the horizontal force applied is 4987N

8 0

3 years ago

At the equator, near the surface of the Earth, the magnetic field is approximately 50.0 μT northward, and the electric field is

n200080 [17]

a) $8.93\cdot 10^{-30} N$ , downward

b) $1.76\cdot 10^{-17} N$ , upward

c) $1.20\cdot 10^{-17} N$ , downward

Explanation:

The gravitational force of an object (also known as weight of the object) is the attractive force with which the object is pulled towards the Earth's centre.

For an object near the Earth's surface, the magnitude of the gravitational force is given by the equation

$F=mg$

where

m is the mass of the object

g is the acceleration due to gravity

In this problem, we have:

$m=9.11\cdot 10^{-31} kg$ is the mass of the electron

$g=9.8 m/s^2$ is the acceleration due to gravity

Therefore, the gravitational force on the electron is:

$F=(9.11\cdot 10^{-31})(9.8)=8.93\cdot 10^{-30} N$

And the direction is downward, towards the Earth's centre.

The electric force on a charged particle is the force produced by the presence of an electric field.

In particular, this force is:

- Repulsive (away from the source of the field) if the charge has the same sign of the charge source of the field

- Attractive (towards the source of the field) if the charge has opposite sign to the charge source of the field

The magnitude of the electric force is given by:

$F=qE$

where

q is the charge of the particle

E is the strength of the electric field

In this problem:

$q=1.6\cdot 10^{-19} C$ is the charge of the electron

$E=110 N/C$ is the strength of the electric field

Substituting,

$F=(1.6\cdot 10^{-19})(110)=1.76\cdot 10^{-17} N$

And since the electron has negative charge, the direction of the force is opposite to that of the electric field, so upward.

When a charged particle is moving in a magnetic field, it experiences a force which is perpendicular to both the direction of motion of the charge and to the direction of the magnetic field.

The magnitude of this force is given by (if the charge moves perpendicular to the magnetic field)

$F=qvB$