What is the acceleration of the car at segment C?

An electric motor has an effective resistance of 22 22 and an inductive reactance of 72 12 when working under load. The voltage

Alecsey [184]

Answer:

Current amplitude, I = 5.57 A

Explanation:

It is given that,

Effective resistance, R = 22 ohms

Inductive reactance, L = 72 ohms

Voltage of the alternating source, V = 420 volt

The total impedance of the RL circuit is given by :

$Z=\sqrt{R^2+L^2}$

$Z=\sqrt{(22)^2+(72)^2}$

Z = 75.28 ohms

From Ohm's law,

$V=I\times Z$

$I=\dfrac{V}{Z}$

$I=\dfrac{420}{75.28}$

I = 5.57 A

So, the current amplitude of the electric motor is 5.57 A. Hence, this is the required solution.

6 0

4 years ago

Particles q1 = -53.0 uc, q2 = +105 uc, and

Nimfa-mama [501]

Answer:

$-180.38\ \text{N}$

Explanation:

$q_1=-53\ \mu\text{C}$

$q_2=105\ \mu\text{C}$

$q_3=-88\ \mu\text{C}$

r = Distance between the charges

$r_{12}=0.5\ \text{m}$

$r_{23}=0.95\ \text{m}$

$r_{13}=1.45\ \text{m}$

k = Coulomb constant = $9\times 10^9\ \text{Nm}^2/\text{C}^2$

Net force is given by

$F=F_{12}+F_{13}\\\Rightarrow F=\dfrac{kq_1q_2}{r_{12}^2}+\dfrac{kq_1q_3}{r_{13}^2}\\\Rightarrow F=kq_1(\dfrac{q_2}{r_{12}^2}+\dfrac{q_3}{r_{13}^2})\\\Rightarrow F=9\times 10^9\times (-53\times 10^{-6})(\dfrac{105\times 10^{-6}}{0.5^2}+\dfrac{-88\times 10^{-6}}{1.45^2})\\\Rightarrow F=-180.38\ \text{N}$