Aluminum foil reflects more light
Complete question is;. A 73mH solenoid inductor is wound on a form that is 0.80m long and 0.10m in diameter a coil having a resistance of 7.7 ohms is tightly wound around the solenoid at its center the mutual inductance of the coil and solenoid is 19μH at a given instant the current in the solenoid is 820mA and is decreasing at the rate of 2.5A/s at the given instant what is the induced current in the coil
Answer:
6.169 μA
Explanation:
Formula for induced EMF is given by the equation;
EMF = M(di/dt). We are given;
di/dt = 2.5 A/s
M = 19μH = 19 × 10^(-6) H
Thus;
EMF = 19 × 10^(-6) × 2.5.
EMF = 47.5 × 10^(-6) V
Formula for current is;
i = EMF/R. R is resistance given as 7.7 ohms.
Thus; i = 47.5 × 10^(-6)/7.7
i = 6.169 μA
Answer:
F = (913.14 , 274.87 )
|F| = 953.61 direction 16.71°
Explanation:
To calculate the resultant force you take into account both x and y component of the implied forces:
Thus, the net force over the body is:
Next, you calculate the magnitude of the force:
and the direction is:
Answer:
yes, should be
Explanation:
This is a hard yes or no question becuase the amplitudes are the same height but in different beating orders.
