Answer:
Explanation:
Given that,
Radius r = 15cm = 0.15m
Area of the circular loop can be determined using the formula for area of a circle
A = π r²
A = π × 0.15²
A = 0.0708 m²
Magnetic field B = 1.2T in positive z direction
B = 1.2 •k T.
If loop is remove from the field in the time interval
∆t = 2.3ms = 2.3×10^-3s
We want to find the average EMF and it is given as
ε = —∆Φ/∆t
The final flux is zero
Φf = 0
Where magnetic flux is given as
Φi = BA Cosθ
Where θ=0 since the area and the magnetic field point in the same direction.
Φi = BA Cos0
Φi = BA
Φi = 1.2 × 0.0708
Φi = 0.0848 Vs
Then, ε = —∆Φ/∆t
ε = —(Φf — Φi) / ∆t
ε = —(0-0.0848) / (2.3×10^-3)
ε = 0.0848 / (2.3×10^-3)
ε = 36.88 V
The EMF is 36.88 Volts
Explanation:
Given that,
The voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10⁻⁸ C.
(a) Capacitance of capacitor 1,

Capacitance of capacitor 2,

Capacitance of capacitor 3,

(b) The equivalent capacitance in series combination is :

Hence, this is the required solution.
Answer:
B
Explanation:
The impulse experienced by an object is the force•time.
Answer:

Explanation:
Since work done is in the form of potential energy, we will use the formula of potential energy here.
We know that,
<h3>P.E. = mgh </h3>
Where,
m = mass = 20 kg
g = acceleration due to gravity = 10 m/s²
h = vertical height = 20 m
So,
<h3>Work done = mgh</h3>
Work done = (20)(10)(20)
Work done = 4000 joules
Work done = 4 kJ
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