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Inga [223]
3 years ago
14

Which best compares the gravitational force and the strong force ?

Physics
2 answers:
Elan Coil [88]3 years ago
8 0
<span>The heavier the body is, the stronger its gravitational pull. Example, the Milky Way Galaxy has a gravitational pull because of the heavenly bodies such as stars and planets are surrounding it. A strong force is exerted if the mass of another body is bigger than the other body.</span>
Natali [406]3 years ago
3 0

Explanation:

Gravitational force :

Gravitation force is an attractive force which act between two bodies.

This force acts between the objects with mass.

The gravitational force is the proportional to the product of their masses of the bodies and inversely proportional to the square of distance between the bodies.

Gravitational force only attraction force.

The gravitational force is very weak force.

Strong force :

The force which holds nucleus together it called strong force.

This force is attractive and repulsive force.

The strong force like electromagnetic force it has short range.

The range of strong force is 10⁻¹⁵ m.

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Typical Pressurized Water Reactors can produce 1100 to 1500
lozanna [386]

Answer: about 1,100,000,000 to 1,500,000,000 Joules/second

Explanation:

1 MW (megawatt) = 1,000,000.00 J/s (joules per second)

1100(1,000,000) = 1,100,000,000

1500(1,000,000) = 1,500,000,000

3 0
2 years ago
Two ball bearings of mass m each moving in opposite directins with equal speed v collide head on with each other.predict the out
posledela

Yo sup??

since the collision is elastic therefore we can say that the two balls will then move in opposite direction.

If ball 1 was moving from east to west then after collision it will move from west to east

and if ball 2 was moving from west to east then it will start moving from east to west.

Hope this helps.

7 0
3 years ago
A measure of randomness or disorder
vodka [1.7K]
It is simply called Entropy.

3 0
3 years ago
Read 2 more answers
3. A football is kicked with a speed of 35 m/s at an angle of 40°.
jarptica [38.1K]

a) 22.5 m/s

The initial vertical velocity is given by:

u_y = u sin \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_y = (35)(sin 40)=22.5 m/s

b) 26.8 m/s

The initial horizontal velocity is given by:

u_x = u cos \theta

where

u = 35 m/s is the initial speed

\theta=40^{\circ} is the angle of projection of the ball

Substituting into the equation, we find

u_x = (35)(cos 40)=26.8 m/s

c) 2.30 s

The time it takes for the ball to reach the maximum heigth can be found by considering the vertical motion only. This is a uniformly accelerated motion (free-fall), so we can use the suvat equation

v_y = u_y + at

where

v_y is the vertical velocity at time t

u_y = 22.5 m/s

a=g=-9.8 m/s^2 is the acceleration of gravity (negative because it is downward)

At the maximum height, the vertical velocity becomes zero, v_y =0; substituting, we find the time t at which this happens:

0=u_y + gt\\t=-\frac{u_y}{g}=-\frac{22.5}{-9.8}=2.30 s

d) 25.8 m

The maximum height can also be found by considering the vertical motion only. We can use the following suvat equation:

s=u_y t + \frac{1}{2}gt^2

where

s is the vertical displacement at time t

u_y = 22.5 m/s

g=-9.8 m/s^2

Substituting t = 2.30 s, we find the displacement at maximum height, so the maximum height:

s=(22.5)(2.30)+\frac{1}{2}(-9.8)(2.30)^2=25.8 m

e) 123.3 m

In order to find how far does the ball lands, we have to consider the horizontal motion.

First of all, the time it takes for the ball to go back to the ground is twice the time needed for reaching the maximum height:

t=2(2.30 s)=4.60 s

Then, we consider the horizontal motion. There is no acceleration along this direction, so the horizontal velocity is constant:

v_x = 26.8 m/s

Therefore, the horizontal distance travelled during the whole motion is

d=v_x t = (26.8)(4.60)=123.3 m

So, the ball lands 123.3 m far from the initial point.

4 0
3 years ago
What is the unique geological feature found on Mercury surface?
Cloud [144]

Answer:

The surface of Mercury has landforms that indicate its crust may have contracted. They are long, sinuous cliffs called lobate scarps. These scarps appear to be the surface expression of thrust faults, where the crust is broken along an inclined plane and pushed upward.

Explanation:

I hope this helps a little bit.

3 0
3 years ago
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