Answer:
aaksj
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
Option (a) is correct.
Falling objects accelerate as they approach the ground.This is because of the force of gravity acting on the falling objects. so the velocity of these objects increases continuously as they approach the ground. the acceleration acting on the falling objects is a constant ( close to the surface of earth) and is called as acceleration due to gravity denoted by g. value of g=9.8 m/s².
Ok. PEMDAS tells us to take care of the square first. When we do that, the denominator becomes
(6.4)^2 x 10^12
= 40.96 x 10^12 .
Now it's just a matter of mashing out the fraction.
The 'mantissa' (the number part) is
6/40.96 = 0.1465
and the order of magnitude is
10^24 / 10^12 = 10^12 .
Put it all together and you've got
1.465 x 10^11 .