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3 years ago

If gravity did not affect the pain of a horizontally thrown ball the ball would

2 answers:

8 0

If gravity did not affect a horizontally thrown ball, the ball would continue along its horizontal path indefinitely (assuming it is thrown in a vacuum so no other forces are acting on it).

Drupady [299]3 years ago

4 0

If gravity had no effect on a ball after you threw it ... and there also
were no air to slow it down ... then the ball would continue traveling
in a straight line, in whatever direction you threw it.

That's the heart and soul of Newton's laws of motion ... any object
keeps moving at the same speed, and in a straight line in the same
direction, until a force acts on it to change its speed or direction.\

If you threw the ball horizontally, then it would keep moving in the
same direction you threw it. But don't forget: The Earth is not flat.
The Earth is a sphere. So, as the ball kept going farther and farther
in the same straight line, the Earth would curve away from it, and it
would look like the ball is getting farther and farther from the ground.

You might be interested in

It is known that the gravitational force of attraction between two alpha particles is much weaker than the electrical repulsion.

natali 33 [55]

Answer:

The ratio of gravitational force to electrical force is 3.19 x 10^-36

Explanation:

mass of an alpha particle = 6.64 x $10^{-27}$ kg

charge on an alpha particle = +2e = +2(1.6 x $10^{-19}$ C) = 3.2 x $10^{-19}$ C

distance between particles = d

For gravitational attraction:

The force of gravitational attraction F = $\frac{Gm^{2} }{r^{2} }$

where G = gravitational constant = 6.67 x $10^{-11}$ m^3 kg^-1 s^-2

r = the distance between the particles = d

m = the mass of each particle

therefore, gravitational force = $\frac{6.67*10^{-11}*(6.64*10^{-27} )^{2} }{d^{2} }$ = $\frac{2.94*10^{-63} }{d^{2} }$ Newton

For electrical repulsion:

Electrical force between the particles = $\frac{-kQ^{2} }{r^{2} }$

where k is the Coulomb's constant = 9.0 x $10^{9}$ N•m^2/C^2

r = distance between the particles = d

Q = charge on each particle

therefore, electrical force = $\frac{-9*10^{9}*(3.2*10^{-19} )^{2} }{d^{2} }$ = $\frac{-9.216*10^{-28} }{d^{2} }$ Newton

the negative sign implies that there is a repulsion on the particles due to their like charges.

Ratio of the magnitude of gravitation to electrical force = $\frac{2.94*10^{-63} }{9.216*10^{-28} }$

==> 3.19 x 10^-36

8 0

3 years ago

The diagram shows what happens to a system undergoing an adiabatic process.

posledela

The answer is:
B. X: Work is done to the system and temperature increases.
Y: Work is done by the system and temperature decreases.

5 0

3 years ago

Read 2 more answers

A tennis ball is dropped from a height of 3 m and bounces back to a height of

julsineya [31]

Answer:

To decide where the balls land, we need to determine how long the balls are in the air. Both balls will take 2 seconds to hit the ground.

Explanation:

1) Time played forward : gravity & drag forces are in opposite directions so it takes a longer time to reach the ground. 2) Time played backward : gravity & drag forces are in the same direction so it takes a shorter time to reach the ground.

5 0

2 years ago

A particle (charge = +0.8 mC) moving in a region where only electric forces act on it has a kinetic energy of 6.7 J at point A.

Maksim231197 [3]

Answer:

The kinetic energy of the particle as it moves through point B is 7.9 J.

Explanation:

The kinetic energy of the particle is:

$\Delta K = \Delta E_{p} = q\Delta V$

Where:

K: is the kinetic energy

$E_{p}$ : is the potential energy

q: is the particle's charge = 0.8 mC

ΔV: is the electric potential = 1.5 kV

$\Delta K = q \Delta V= 0.8 \cdot 10^{-3} C*1.5 \cdot 10^{3} V = 1.2 J$

Now, the kinetic energy of the particle as it moves through point B is:

$\Delta K = K_{f} - K_{i}$

$K_{f} = \Delta K + K_{i} = 1.2 J + 6.7 J = 7.9 J$

Therefore, the kinetic energy of the particle as it moves through point B is 7.9 J.

I hope it helps you!

8 0

3 years ago

You drive at 195 miles down the 101 freeway at 65 miles per hour. How long did it take? time =

trasher [3.6K]

Answer:

According to my calculations, if you drive an average of 65 miles per hour, with 0 minutes of stop-time, you should reach your destination in 3 hours and 0 minutes.

Explanation:

7 0

3 years ago