b. What speed must the student leave the ground with to reach that height?
The hang time of the student is 0.64 seconds, and he must leave the ground with a speed of 3.13 m/s
Why?
To solve the problem, we must consider the vertical height reached by the student as max height.
We can use the following equations to solve the problem:
Initial speed calculations:
At max height, the speed tends to zero.
So, calculating, we have:
Hang time calculations:
We must remember that the total hang time is equal to the time going up plus the time going down, and both of them are equal,so, calculating the time going down, we have have:
Then, for the total hang time, we have:
Have a nice day!
Answer: The diameter of the circular path is 2.96m
Explanation: centripetal acceleration = tangential speed^2 / radius of the circular path.
Centripetal acceleration = 2.7m/s^2
Tangential speed = 2.0m/s
Radius = 2.0^2 / 2.7 = 4/2.7
= 1.48m
Diameter = radius*2
= 1.48*2 = 2.96m.
Answer:
If it continues long enough, then the object
returns to where it started.
Explanation:
By definition, a region of space that contains no matter at all is a vacuum.
They add energy to the work.
B.
They increase the force needed to work.
C.
They reduce the speed of work.
D.
They increase the distance over which a force is applied.
C.?
Answer: D. They increase the distance over which a force is applied.
Explanation:
Simple machines are basic mechanical devices which make our work easier like lever, pulley, wheel and axle, inclined plane, wedge and screw. The magnitude of the force or direction is altered using a simple machine. They increase the distance over which a force is applied. They use mechanical advantage to increase the magnitude of the force.