To be able to write an ss-domain equation for a circuit, use partial fraction decomposition to separate the terms in this equati
on, and then inverse transform the equation back into the time-domain. Using the mesh-current method to analyze the following circuit yields a set of two integro-differential equations which would be difficult to solve analytically. However, applying a Laplace transform allows for these equations to be rewritten as polynomial equations which can be manipulated algebraically.
1 answer:
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Answer:
Amperes WB =<em> </em>
Explanation:
Amperes will flow back to the negative terminal because both the positive terminal and negative terminal contain the equivalent amount of current that will flow and it works in the starter motor.
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The load is 17156 N.
<u>Explanation:</u>
First compute the flexural strength from:
σ = FL / π
= 3000 (40
10^-3) / π (5
10^-3)^3
σ = 305 10^6 N / m^2.
We can now determine the load using:
F = 2σd^3 / 3L
= 2(305 10^6) (15
10^-3)^3 / 3(40
10^-3)
F = 17156 N.
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