Given:

frequency, f = 60.0 Hz
frequency, f' = 45.0 Hz

Solution:
To calculate max current in inductor,
:
At f = 60.0 Hz


L = 0.1326 H
Now, reactance
at f' = 45.0 Hz:


Now,
is given by:
Therefore, max current in the inductor,
= 2.13 A
Answer:
T = 15 kN
F = 23.33 kN
Explanation:
Given the data in the question,
We apply the impulse momentum principle on the total system,
mv₁ + ∑
= mv₂
we substitute
[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ × ( 45 × 1000 / 3600 )
F( 75 - 0 ) = 1.75 × 10⁶
The resultant frictional tractive force F is will then be;
F = 1.75 × 10⁶ / 75
F = 23333.33 N
F = 23.33 kN
Applying the impulse momentum principle on the three cars;
mv₁ + ∑
= mv₂
[3(30)]×10³ × 0 + FΔt = [3(30)]×10³ × ( 45 × 1000 / 3600 )
F(75-0) = 1.125 × 10⁶
The force T developed is then;
T = 1.125 × 10⁶ / 75
T = 15000 N
T = 15 kN
54 volts
Ohms law. E= I x R
Answer:

Explanation:
We have to combine the following formula to find the mass yield:


The diffusion coefficient : 
The area : 
Time : 
ΔC: 
Δx: 
Now substitute the values

![M=-(6.0*10^{-8} m/s^{2})(0.25 m^{2})(3600 s/h)[(0.64-3.0kg/m^{3})(3.1*10^{-3}m)]](https://tex.z-dn.net/?f=M%3D-%286.0%2A10%5E%7B-8%7D%20m%2Fs%5E%7B2%7D%29%280.25%20m%5E%7B2%7D%29%283600%20s%2Fh%29%5B%280.64-3.0kg%2Fm%5E%7B3%7D%29%283.1%2A10%5E%7B-3%7Dm%29%5D)

Capillary action occurs When the adhesion to the walls stronger than dirt cohesive forces between a liquid molecules. the head towards Capillery action will take water in a uniform circular is limited by surface tension and, of course, gravity.