A 2.44 kg block is pushed 1.55 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle o
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Answer:
induced EMF = 240 V
and by the lenz's law direction of induced EMF is opposite to the applied EMF
Explanation:
given data
inductance = 8 mH
resistance = 5 Ω
current = 4.0 A
time t = 0
current grow = 4.0 A to 10.0 A
to find out
value and the direction of the induced EMF
solution
we get here induced EMF of induction is express as
E = - L ...................1
so E = - L
put here value we get
E = - 8 ×
E = -40 × 6
E = -240
take magnitude
induced EMF = 240 V
and by the lenz's law we get direction of induced EMF is opposite to the applied EMF
Answer:
d= 794.4 cmExplanation:
Given that
Speed ,V= 286 km/h
V=79.44 m/s
Given that time ,t= 100 ms
t= 0.1 s
We know that ( if acceleration is zero)
Distance = Speed x time
d= V t
Now by putting the values in the above equation
d = 79.44 x 0.1 m
d= 7.944 m
We know that 1 m = 100 cm
d= 794.4 cm