Answer:
d. 332 V
Explanation:
Given;
number of turns in the wire, N = 40 turns
area of the coil, A = 0.06 m²
magnitude of the magnetic field, B = 0.4 T
frequency of the wave, f = 55 Hz
The maximum emf induced in the coil is given by;
E = NBAω
Where;
ω is angular velocity = 2πf
E = NBA(2πf)
E = 40 x 0.4 x 0.06 x (2 x π x 55)
E = 332 V
Therefore, the maximum induced emf in the coil is 332 V.
The correct option is "D"
d. 332 V
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:
a) It takes her 1.43 s to reach a speed of 2.00 m/s.
b) Her deceleration is - 2.50 m/s²
Explanation:
The equation of velocity for an object that moves in a straight line with constant acceleration is as follows:
v = v0 + a · t
Where:
v = velocty.
v0 = initial velocity.
a = acceleration.
t = time.
a) Using the equation of velocity, let´s consider that the car moves in the positive direction. Then:
v = v0 + a · t
2.00 m/s = 0 m/s + 1.40 m/s² · t
t = 2.00 m/s / 1.40 m/s²
t = 1.43 s
It takes her 1.43 s to reach a speed of 2.00 m/s
b) Let´s use again the equation of velocity, knowing that at t = 0.800 s the velocity is 0 m/s:
v = v0 + a · t
0 = 2.00 m/s + a · 0.800 s
-2.00 m/s / 0.800 s = a
a = -2.50 m/s²
Her deceleration is - 2.50 m/s²
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