The second answer choice is not true (ozone is concentrated in the stratosphere, not the exosphere)
First, you must know that the statement "<span>An object at rest tends to stay at rest. An object in motion tends to stay in motion unless acted upon by an outside force." is true, because it is the first Law of Newton or inercy law.
what outside force acts on a baseball when it is thrown straight like a pitcher pitching a ball?
After the ball leaves the pitcher's hand, it is subject only to the gravitational attraction of the Earth. That is why the pitcher has to give the appropiate impulse in order to the ball reaches the point that he and the catcher want.
What about if you threw it straight into the air?
It is the same thing. The only force would be the gravitational attraction of the Earth.
What about if you threw the baseball in outer space. Would there be any forces to slow that down?
In outer space, at the beginning the baseball would be very far from of other massive objects to feel their gravitational field, so there would not be any forces to slow it down. Although eventulally, after many light-years, it would enter the gravitational field of a galaxy or other massive body and it would attract it.
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The force applied to the second ball by the first ball is 6.734 × 10^-4 N.
<h3>What is impulse of force?</h3>
The impulse of force is defined as the sum of the average force and the duration it is applied.
If the mass of the item remains constant, the impulse of force equals the change in momentum of the object.
Given that: mass of a metal sphere: m = 0.026 kg.
Initial speed of the sphere: u = 3.7 m/s.
When the sphere stops completely, its change in momentum = mu - 0
= 0.026×3.7 N-s.
= 0.0962 N-s.
As the spheres are in contact for 0.007s before the second sphere is shot off down the track, the force applied to the second ball =
change in momentum of 1st ball × time of contact
= 0.0962 × 0.007 N
= 0.0006734 N
= 6.734 × 10^-4 N.
Hence, the force applied to the second ball is 6.734 × 10^-4 N.
Learn more about impulse force here:
brainly.com/question/29787329
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The distance to the water is the same for both ... call it s meters
They both take 1.6 s to reach the water ... so t = 1.6 seconds
They both step off the bridge ... so both have an initial vertical velocity is 0 m/s
Just consider the vertical motion and take DOWN as the positive direction
a = 9.8 m/s²
s = v(i)t + (1/2)at²
s = 0 + (1/2) * 9.8 * 1.6²
s = 12.5 m ←←← Edit: Forgot to say ... that's how high the bridge is above the water
Now get the time it takes the first jumper to reach 1.8 m:
s = v(i)t + (1/2)at²
1.8 = 0 + 4.9t²
t = 0.61 s
so when the 2nd person jumps it takes the first person another 1.6 - 0.61 = 0.99 s to reach the water
Now find how far the 2nd jumper falls in 0.99 s:
s = 0 + 4.9 * 0.99²
s = 4.8 m
so the separation distance between the two jumpers when the 1st jumper hits the water is 12.5 - 4.8 = 7.7 m
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