A student plans an investigation to determine whether an object has a positive charge, a negative charge, or is electrically neu
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.
This question is incomplete, the complete question is;
In a softball windmill pitch, the pitcher rotates her arm through just over half a circle, bringing the ball from a point above her shoulder and slightly forward to a release point below her shoulder and slightly forward. (Figure 1) shows smoothed data for the angular velocity of the upper arm of a college softball pitcher doing a windmill pitch; at time t = 0 her arm is vertical and already in motion. For the first 0.15 s there is a steady increase in speed, leading to a final push with a greater acceleration during the final 0.05 s before the release. In each part of the problem, determine the corresponding quantity during the first 0.15 s of the pitch.
Angular Velocity at time 0s = 12 rad/s
Angular Velocity at time 0.15s = 24 rad/s
a) What is the angular acceleration?
b) If the ball is 0.60 m from her shoulder, what is the tangential acceleration of the ball? This is the key quantity here--it's a measure of how much the ball is speeding up. Express your answer in m/s2 and in units of g
Answer:
a) the angular acceleration is 80 rad/s²
b) the tangential acceleration of the ball is;
- a = 48 m/s²
- a = 4.9 g
Explanation:
Given the data in the question;
from the graph below;
Angular Velocity at time 0s = 12 rad/s
Angular Velocity at time 0.15s = 24 rad/s
a) What is the angular acceleration;
Angular acceleration ∝ = ( -
) / dt
we substitute
Angular acceleration ∝ = ( 24 - 12 ) / 0.15
Angular acceleration ∝ = 12 / 0.15
Angular acceleration ∝ = 80 rad/s²
Therefore, the angular acceleration is 80 rad/s²
b)
If the ball is 0.60 m from her shoulder, i.e s = 0.6 m
the tangential acceleration of the ball will be;
a = ∝ × s
we substitute
a = 80 × 0.6
a = 48 m/s²
a = ( 48 / 9.8 )g
a = 4.9 g
Therefore, the tangential acceleration of the ball is;
- a = 48 m/s²
- a = 4.9 g
