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Tom [10]
3 years ago
14

If gravity between the sun and Earth suddenly vanished; describe the expected Earth motion.

Physics
1 answer:
Daniel [21]3 years ago
6 0

Answer:

When the gravity between the Sun and the Earth suddenly vanishes, then Earth will keep moving in a straight line in a direction where it was moving at the moment when gravity vanished.

Explanation:

The gravitational force between Sun and Earth is given by

F = G\frac{M_{1}M_{2} }{R^{2}}

Where F is the gravitational force between the Sun and the Earth. M₁ and M₂ are the masses of Sun and Earth and R is the distance between them.

If we assume that the gravity between the sun and Earth suddenly vanishes, then there would not be any force between the Sun and the Earth and the Earth will keep on moving in a straight line. This is endorsed by the Newton's first law that a body in motion remains in motion if no external force is acting on it.

The direction of Earth's motion will be determined by the previous direction of the motion that is the moment when the gravity was vanished.

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A ball is thrown vertically upwards with an initial velocity of 20.00 m/s. Neglecting air resistance, how long is the ball in th
algol [13]

Answer:

1) The greatest height attained by the ball equals 20.387 meters.

2) The time it takes for the ball to reach 15 meters approximately equals 1 second.

Explanation:

The greatest height will be attained when the ball stop's in the air and starts falling back to the earth.

thus using third equation of kinematics we obtain the height attained as

v^2=u^2+2as

where

'v' is the final speed of the ball

'u' is the initial speed of the ball

'a' is the acceleration that the ball is under which in this case equals 9.81 m/s^{2}

's' is the distance it covers

Thus for maximum height applying the values in the equation we get

0=20^{2}-2\times 9.81\times h\\\\\therefore h=\frac{20^{2}}{2\times 9.81}=20.387meters

Using the same equation we can find the speed of the ball when it reaches 15 meters of height as  

v^2=20^{2}-2\times 9.81\times 15\\\\v^{2}=105.7\\\\\therefore v=10.28m/s

the time it takes to reduce the velocity to this value can be found by first equation of kinematics as

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4 0
4 years ago
A 0.0250-kg bullet is accelerated from rest to a speed of 550 m/s in a 3.00-kg rifle. The pain of the rifle’s kick is much worse
kondaur [170]

Answer:

a) 4.583 m/s

b) 31.505 J

c) 0.491 m/s

d) 3.375 J

e)

   p_player = (110 kg)(8 m/s) = 880 kg m/s

   p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

Explanation:

HI!

a)

We can calculate the recoil velocity by conservation of momentum, remember that p=mv.

The momentum of the bullet is:

p_b = (0.0250 kg)*(550 m/s )

The momentum of the rifle is:

p_r = (3 kg) * v

Since the total initial momentum is zero:

p_b = p_r

That is:

v = (550 m/s ) (0.0250 kg/ 3 kg ) = 4.583 m/s

b)

The kinetic energy gained by the rifle is:

K = (1/2) m v^2 = (1/2) *(3 kg) *(4.583 m/s)^2 = 31.505 J

c)

We use the same formula as in a), but with m=28kg instead of 3 kg

v = (550 m/s ) (0.0250 kg/ 28 kg ) = 0.491 m/s

d)

Again, the same formula as b, but with m=28 and v=0.491 m/s

K = 3.375 J

e)

p_player = (110 kg)(8 m/s) = 880 kg m/s

p_ball = (0.41 kg)(25 m/s) = 10.25 kg m/s

I believe that the kinetic energy is more related to the problem than the momentum. The relation between these two quantities is:

K = p^2/(2m)

usiing this relation, we get:

K_player = 3520 J

K_ball =  128.125 J

Therefore the kinetic energy of the player is around 27 time larger than the kinetic energy of the ball, that being said, the pain of being tackled by that player is around 27 times greater that being hit by the ball!

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Explanation:

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