Answer:
The induced current and the power dissipated through the resistor are 0.5 mA and
.
Explanation:
Given that,
Distance = 1.0 m
Resistance = 3.0 Ω
Speed = 35 m/s
Angle = 53°
Magnetic field 
(a). We need to calculate the induced emf
Using formula of emf

Where, B = magnetic field
l = length
v = velocity
Put the value into the formula


We need to calculate the induced current


Put the value into the formula


(b). We need to calculate the power dissipated through the resistor
Using formula of power

Put the value into the formula


Hence, The induced current and the power dissipated through the resistor are 0.5 mA and
.
The total displacement is equal to the total distance. For the east or E direction, the distance is determined using the equation:
d = vt = (22 m/s)(12 s) = 264 m
For the west or W direction, we use the equations:
a = (v - v₀)/t
d = v₀t + 0.5at²
Because the object slows down, the acceleration is negative. So,
-1.2 m/s² = (0 m/s - 22 m/s)/t
t = 18.33 seconds
d = (22 m/s)(18.33 s) + 0.5(-1.2 m/s²)(18.33 s)²
d = 201.67 m
Thus,
Total Displacement = 264 m + 201.67 m = 465.67 or approximately 4.7×10² m.
Acceleration is measured in m/s².
Answer: m/s²
The charge of the object must be 
Answer: Option C
<u>Explanation:</u>
Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.
Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.

Here, given E = 4500 N/C and F = 0.05 N.
We need to find charge of the object (q)
By substituting the given values, we get

Answer:
Explanation:
Single-phase transformers can operate to either increasing or decreasing the voltage applied to the primary winding. When a transformer is used to “increase” the voltage on the secondary winding with respect to the primary, it is called a Step-up transformer