Answer:
The final velocity of the thrower is and the final velocity of the catcher is .
Explanation:
Given:
The mass of the thrower, .
The mass of the catcher, .
The mass of the ball, .
Initial velocity of the thrower,
Final velocity of the ball,
Initial velocity of the catcher,
Consider that the final velocity of the thrower is . From the conservation of momentum,
Consider that the final velocity of the catcher is . From the conservation of momentum,
Thus, the final velocity of thrower is and that for the catcher is .
Answer:
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Explanation:
According to this problem, the rocket is accelerated uniformly due to thrust during 30 seconds and after that is decelerated due to gravity. The velocity as function of initial velocity, acceleration and time is:
(1)
Where:
- Initial velocity, measured in meters per second.
- Final velocity, measured in meters per second.
- Acceleration, measured in meters per square second.
- Initial time, measured in seconds.
- Final time, measured in seconds.
Now we obtain the kinematic equations for thrust and free fall stages:
Thrust (, , , )
(2)
Free fall (, , , )
(3)
Now we created the graph speed-time, which can be seen below.
The CNS (Central Nervous System) consists of the brain and spinal cord.
Answer:
Amplitude = 0.058m
Frequency = 6.25Hz
Explanation:
Given
Amplitude (A) = 8.26 x 10-2 m
Frequency (f) = 4.42Hz
Conversation of energy before split
½mv² = ½KA²
Make A the subject of formula
A =
Conversation of energy after split
½(m/2)V'² = ½(m/2)V² = ½KA'²
½(m/2)V² = ½KA'²
Make A the subject of formula
First divide both sides by ½
(m/2)V² = KA'²
Divide both sides by K
V² = A'²
= A'
Substitute for A in the above equation
A' = A/√2
A' = 8.26 x 10^-2/√2
A' = 0.05840702012600882
Amplitude after split = 0.058 (Approximated)
Frequency (f') = f√2
f' = 4.42√2
f' = 6.25082394568908011
Frequency after split = 6.25Hz (approximated)
The sun is a clear example of objects releasing radiation in nature