"Given the acceleration of an object as a function of time, what additional information is necessary to use integration to deter
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Answer:
- After throwing the snow, velocity of the thrower is 2.33 m/s
- the velocity of the receiver is 0.026 m/s
Explanation:
Given the data in the question;
Using conservation of momentum,
Initial thrower has a momentum of mv; v
(69.5 kg + 0.0475 kg) × 2.35 m/s = 163.4366 kg.m/s
Now, When he throws it at 31.5 m/s, these constitutes a momentum of;
(0.0475 kg )(31.5 m/s) = 1.49625 kg.m/s
hence his momentum now is: 163.4366 - 1.49625 = 161.94035 kg.m/s
To get his velocity, we say;
161.94035 = mv
{ he lost weight of the snow ball so, m = 69.5 kg )
161.94035 = 69.5 × v
v = 161.94035 / 69.5
v = 2.33 m/s
Therefore, After throwing the snow, velocity of the thrower is 2.33 m/s
Next is the Receiver;
the receiver will gain momentum of 1.49625 kg.m/s
he has no momentum initially and after he catches the snow ball;
1.49625 kg.m/s = mv
1.49625 kg.m/s = ( 57.5 kg + 0.0475 kg ) × v
1.49625 kg.m/s = 57.5475 kg × v
v = ( 1.49625 kg.m/s ) / 57.5475 kg
v = 0.026 m/s
Therefore, the velocity of the receiver is 0.026 m/s
Hello
The bullet is moving by uniformly accelerated motion.
The initial velocity is
, the final velocity is 
, and the total time of the motion is 
, so the acceleration is given by
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using
So, the bullet penetrates the sandbag 1.8 meters.
The bullet is moving by uniformly accelerated motion.
The initial velocity is
where the negative sign means that is a deceleration.
Therefore we can calculate the total distance covered by the bullet in its motion using
So, the bullet penetrates the sandbag 1.8 meters.