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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:
I believe it’s self-referent encoding
Hey there mate :)
Even if two persons are given the same work load, the speed of the work done gets different by the energy of those persons.
No one is sure that he/she can complete the work within the time. He may or may not.
Also, the physical characteristics makes the work different. If one person has so much power to work all day, the other person may not have.
Therefore, <em>even if two persons do the same amount of work , they may have different power</em><em>.</em>
I think B is the most correct, because logically it's harder to bend a stiffer spring than it is to bend a softer one. Also, I don't think length comes into play. So B.
Answer:
Solution:
we have given the equation of motion is x(t)=8sint [where t in seconds and x in centimeter]
Position, velocity and acceleration are all based on the equation of motion.
The equation represents the position. The first derivative gives the velocity and the 2nd derivative gives the acceleration.
x(t)=8sint
x'(t)=8cost
x"(t)=-8sint
now at time t=2pi/3,
position, x(t)=8sin(2pi/3)=4*squart(3)cm.
velocity, x'(t)=8cos(2pi/3)==4cm/s
acceleration, x"(t)==8sin(2pi/3)=-4cm/s^2
so at present the direction is in y-axis.
