Answer
Integral EdA = Q/εo =C*Vc(t)/εo = 3.5e-12*21/εo = 4.74 V∙m <----- A)
Vc(t) = 21(1-e^-t/RC) because an uncharged capacitor is modeled as a short.
ic(t) = (21/120)e^-t/RC -----> ic(0) = 21/120 = 0.175A <----- B)
Q(0.5ns) = CVc(0.5ns) = 2e-12*21*(1-e^-t/RC) = 30.7pC
30.7pC/εo = 3.47 V∙m <----- C)
ic(0.5ns) = 29.7ma <----- D)
Answer:
a) r = 6122 m and b) v = 32.5 m / s
Explanation:
a) The train in the curve is subject to centripetal acceleration
a = v2 / r
Where v is The speed and r the radius of the curve
They indicate that the maximum acceleration of the person is 0.060g,
a = 0.060 g
a = 0.060 9.8
a = 0.588 m /s²
Let's calculate the radius
v = 216 km / h (1000m / 1km) (1 h / 3600 s =
v = 60 m / s
r = v² / a
r = 60² /0.588
r = 6122 m
b) Let's calculate the speed, for a radius curve 1.80 km = 1800 m
v = √a r
v = √( 0.588 1800)
v = 32.5 m / s
Yes, therefore the object would then slow down
<span>Transformed into potential energy</span>
