Answer:
Vi = 94.64 m/s
Explanation:
I order to find out the initial velocity of the object, we can use third equation of motion:
2ah = Vf² - Vi²
where,
a = acceleration = -9.8 m/s²
h = maximum height covered by object = 460 m - 3 m = 457 m
Vf = Final Velocity = 0 m/s (since, object momentarily stops at highest point)
Vi = Initial Velocity = ?
Therefore,
2(-9.8 m/s²)(457 m) = (0 m/s)² - Vi²
Vi = √8957.2 m²/s²
<u>Vi = 94.64 m/s</u>
Answer:
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Explanation:
a) the capacitance is given of a plate capacitor is given by:
C = \epsilon_0*(A/d)
Where \epsilon_0 is a constant that represents the insulator between the plates (in this case air, \epsilon_0 = 8.84*10^(-12) F/m), A is the plate's area and d is the distance between the plates. So we have:
The plates are squares so their area is given by:
A = L^2 = 0.19^2 = 0.0361 m^2
C = 8.84*10^(-12)*(0.0361/0.0077) = 8.84*10^(-12) * 4.6883 = 41.444*10^(-12) F
b) The charge on the plates is given by the product of the capacitance by the voltage applied to it:
Q = C*V = 41.444*10^(-12)*120 = 4973.361 * 10^(-12) C = 4.973 * 10^(-9) C
c) The electric field on a capacitor is given by:
E = Q/(A*\epsilon_0) = [4.973*10^(-9)]/[0.0361*8.84*10^(-12)]
E = [4.973*10^(-9)]/[0.3191*10^(-12)] = 15.58*10^(3) V/m
d) The energy stored on the capacitor is given by:
W = 0.5*(C*V^2) = 0.5*[41.444*10^(-12) * (120)^2] = 298396.8*10^(-12) = 0.298 * 10 ^6 J