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
Answer:
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Answer:
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Explanation:
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Answer:
<em> - 14.943 W/m^2K ( negative sign indicates cooling ) </em>
Explanation:
Given data:
Area of FPC = 4 m^2
temp of water = 60°C
flow rate = 0.06 l/s
ambient temperature = 8°C
exit temperature = 49°C
<u>Calculate the overall heat loss coefficient </u>
Note : heat lost by water = heat loss through convection
m*Cp*dT = h*A * ( T - To )
∴ dT / T - To = h*A / m*Cp ( integrate the relation )
In (
) = h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )
In ( 41 / 52 ) = 0.0159*h
hence h = - 0.2376 / 0.0159
= - 14.943 W/m^2K ( heat loss coefficient )
Answer:
R=1923Ω
Explanation:
Resistivity(R) of copper wire at 20 degrees Celsius is 1.72x10^-8Ωm.
Coil length(L) of the wire=37.0m
Cross-sectional area of the conductor or wire (A) = πr^2
A= π * (2.053/1000)/2=3.31*10^-6
To calculate for the resistance (R):
R=ρ*L/A
R=(1.72*10^8)*(37.0)/(3.31*10^-6)
R=1922.65Ω
Approximately, R=1923Ω