21. A 1000-kg automobile moving with a speed of 24 m/s relative to the road collides with a 500-kg automobile initially at rest.
1 answer:
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Answer:
a) The rocket reaches a maximum height of 737.577 meters.
b) The rocket will come crashing down approximately 17.655 seconds after engine failure.
Explanation:
a) Let suppose that rocket accelerates uniformly in the two stages. First, rocket is accelerates due to engine and second, it is decelerated by gravity.
1st Stage - Engine
Given that initial velocity, acceleration and travelled distance are known, we determine final velocity (), measured in meters per second, by using this kinematic equation:
(1)
Where:
- Acceleration, measured in meters per square second.
- Travelled distance, measured in meters.
- Initial velocity, measured in meters per second.
If we know that ,
and
, the final velocity of the rocket is:
The time associated with this launch (), measured in seconds, is:
2nd Stage - Gravity
The rocket reaches its maximum height when final velocity is zero:
(2)
Where:
- Initial speed, measured in meters per second.
- Final speed, measured in meters per second.
- Gravitational acceleration, measured in meters per square second.
- Initial height, measured in meters.
- Final height, measured in meters.
If we know that ,
,
and
, then the maximum height reached by the rocket is:
The rocket reaches a maximum height of 737.577 meters.
b) The time needed for the rocket to crash down to the launch pad is determined by the following kinematic equation:
(2)
Where:
- Initial height, measured in meters.
- Final height, measured in meters.
- Initial speed, measured in meters per second.
- Gravitational acceleration, measured in meters per square second.
- Time, measured in seconds.
If we know that ,
,
and
, then the time needed by the rocket is:
Then, we solve this polynomial by Quadratic Formula:
,
Only the first root is solution that is physically reasonable. Hence, the rocket will come crashing down approximately 17.655 seconds after engine failure.
Answer:
Explanation:
When a spring is compressed, the force exerted by the spring is given by:
where
k is the spring constant
x is the compression of the spring
In this problem we have:
k = 52 N/m is the spring constant
x = 43 cm = 0.43 m is the compression
Therefore, the force exerted by the spring on the dart is
Now we can apply Newton' second law of motion to calculate the acceleration of the dart:
where
F = 22.4 N is the force exerted on the dart by the spring
m = 75 g = 0.075 kg is the mass of the dart
a is its acceleration
Solving for a,