The position of the particle is given by:
x(t) = t³ - 12t² + 21t - 9
Differentiate x(t) with respect to t to find the velocity x'(t):
x'(t) = 3t² - 24t + 21
Differentiate x'(t) with respect to t to find the acceleration x''(t):
x''(t) = 6t - 24
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<em><u>⇒</u></em>Answer:</h2>
In the standing broad jump, one squats and then pushes off with the legs to see how far one can jump. Suppose the extension of the legs from the crouch position is 0.600 m and the acceleration achieved from this position is 1.25 times the acceleration due to gravity, g . How far can they jump? State your assumptions. (Increased range can be achieved by swinging the arms in the direction of the jump.)
Step-by-Step Solution:
Solution 35PE
This question discusses about the increased range. So, we shall assume that the angle of jumping will be as the horizontal range is maximum at this angle.
Step 1 of 3<
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The legs have an extension of 0.600 m in the crouch position.
So, m
The person is at rest initially, so the initial velocity will be zero.
The acceleration is m/s2
Acceleration m/s2
Let the final velocity be .
Step 2 of 3<
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Substitute the above given values in the kinematic equation ,
m/s
Therefore, the final velocity or jumping speed is m/s
Explanation:
Frequensey or hertz, I looked this up on the internet!
Answer:
C
Explanation:
Velocity is just speed with direction
Answer:
d. equal to one-fourth the acceleration at the surface of the asteroid.
Explanation:
The explanation is attached as a picture with this answer
Newton's law of universal gravitation is being used to compare the accelerations at the surface and at the top of the ball's path.
as it can be seen in the explanation that the proportional form of the equation is used because we do not need to necessarily use to final form with "G" for comparison calculations.
As per the given scenario only difference between the two points in the gravitational field is the distance from center of the spherical asteroid, i.e. r.
It is taken 2r for the top is the path. hence we obtain (1/4)g as our answer.