Block A is released from rest and slides down the frictionless ramp to the loop. The maximum height h of the loop is the same as
the initial height of the block. Will A make it completely around the loop without losing contact with the track?
1 answer:
Answer:No
Explanation:
Given
Block A is at height h and released from rest
Initial Energy possessed by block A is equal to Potential energy of Block which is given by
where m=mass of block
After releasing the block, block first reaches to the bottom of the circle and then uses this energy to reach a top point of the loop.
But as soon as block reaches the top point of the loop it acquires the energy which is equal to Initial energy i.e. all the energy is stored in the form of potential energy and there is no kinetic energy so the block will not able to move further and fall from the top point.
You might be interested in
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Answer:
0
Explanation:
output =transfer function H(s) ×input U(s)
here H(s)=
U(s)= for unit step function
output =H(s)×U(s)
=×
=
taking inverse laplace of output
output=t×
at t=0 putting the value of t=0 in output
output =0
Answer:
Mechanical Efficiency = 83.51%
Explanation:
Given Data:
Pressure difference = ΔP=1.2 Psi
Flow rate =
Power of Pump = 3 hp
Required:
Mechanical Efficiency
Solution:
We will first bring the change the units of given data into SI units.
Now we will find the change in energy.
Since it is mentioned in the statement that change in elevation (potential energy) and change in velocity (Kinetic Energy) are negligible.
Thus change in energy is
As we know that Mass = Volume x density
substituting the value
Energy = Volume * density x ΔP / density
Change in energy = Volumetric flow x ΔP
Change in energy = 0.226 x 8.274 = 1.869 KW
Now mechanical efficiency = change in energy / work done by shaft
Efficiency = 1.869 / 2.238
Efficiency = 0.8351 = 83.51%