GIVEN:
Amplitude, A = 0.1mm
Force, F =1 N
mass of motor, m = 120 kg
operating speed, N = 720 rpm
=
Formula Used:

Solution:
Let Stiffness be denoted by 'K' for each mounting, then for 4 mountings it is 4K
We know that:

so,
= 75.39 rad/s
Using the given formula:
Damping is negligible, so, 
will give the tranfer function
Therefore,
= 
= 
Required stiffness coefficient, K = 173009 N/m = 173.01 N/mm
The ratio between a and b is 1/3
Why 1+12+ Y3 < 1100
Says the state of university Need to purchase 1100 computers in total, we have the following answer on the way top
Answer:
The specific weight of unknown liquid is found to be 15 KN/m³
Explanation:
The total pressure in tank is measured to be 65 KPa in the tank. But, the total pressure will be equal to the sum of pressures due to both oil and unknown liquid.
Total Pressure = Pressure of oil + Pressure of unknown liquid
65 KPa = (Specific Weight of oil)(depth of oil) + (Specific Weight of unknown liquid)(depth of unknown liquid)
65 KN/m² = (8.5 KN/m³)(5 m) + (Specific Weight of Unknown Liquid)(1.5 m)
(Specific Weight of Unknown Liquid)(1.5 m) = 65 KN/m² - 42.5 KN/m²
(Specific Weight of Unknown Liquid) = (22.5 KN/m²)/1.5 m
<u>Specific Weight of Unknown Liquid = 15 KN/m³</u>