Answer:
m 200 g , T 0.250 s,E 2.00 J
;
2 2 25.1 rad s
T 0.250
(a)
2 2
k m 0.200 kg 25.1 rad s 126 N m
(b)
2
2 2 2.00 0.178 mm 200 g , T 0.250 s,E 2.00 J
;
2 2 25.1 rad s
T 0.250
(a)
2 2
k m 0.200 kg 25.1 rad s 126 N m
(b)
2
2 2 2.00 0.178 m
Explanation:
That is a reason
Pretty sure it’s Force*Distance*Cos(theta)
I believe it is called an ampere.
The final velocity (
) of the first astronaut will be greater than the <em>final velocity</em> of the second astronaut (
) to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts <em>after throwing the ball</em>.
The given parameters;
- Mass of the first astronaut, = m₁
- Mass of the second astronaut, = m₂
- Initial velocity of the first astronaut, = v₁
- Initial velocity of the second astronaut, = v₂ > v₁
- Mass of the ball, = m
- Speed of the ball, = u
- Final velocity of the first astronaut, =
- Final velocity of the second astronaut, =
The final velocity of the first astronaut relative to the second astronaut after throwing the ball is determined by applying the principle of conservation of linear momentum.
if v₂ > v₁, then 
, to conserve the linear momentum.
Thus, the final velocity () of the first astronaut will be greater than the <em>final velocity</em> of the second astronaut () to ensure that the total initial momentum of both astronauts is equal to the total final momentum of both astronauts after throwing the ball.
Learn more here: brainly.com/question/24424291
Answer:
<h3>14.97m/s</h3>
Explanation:
Given
Initial velocity of the car u = 8m/s
Distance travelled by the rider S = 40m
Acceleration a = 2m/s²
Required
rider's velocity after the acceleration v
Using the equation of motion
v² = u²+2as
v² = 8²+2(2)(40)
v² = 64+160
v² = 224
v = √224
v = 14.97m/s
Hence the rider's velocity after the acceleration is 14.97m/s
