If you can observe, we are only given one parameter here which is the time. If you want to compute for the distance, you have to know the speed. The hint here is the 'radio transmissions'. All the information gathered by the probe from the space, is sent back to the Earth by electromagnetic waves. Hence, we must know the speed of electromagnetic waves. Since they are as fast as light, their speed is equal to 300 million meters per second. Then, we can finally determine the distance.
d = speed*time
d = (300×10⁶ m/s)(2.53 hours)
Since 1 hour = 3,600 seconds,
d = (300×10⁶ m/s)(2.53 hours)(3,600 seconds/1 hour)
d = 2.73×10⁻¹² m
Answer: 580 x 10^-3 J
Explanation:
0.6mm is 0.6/1000 = 600*10^-6 m
The plate area is .17*.17 = 28.9*10^-3 m^2
Air:
The voltage that can be sustained by 0.60 mm of air dielectric is:
V = 3.0*10^6* 600*10^-6 = 1800 V
The capacitance is:
C = ε*A/d = 8.854*10^-12 * 28.9*10^-3/600*10^-6 = 426*10^-12 F = 426 pF
The energy stored in a capacitor is:
E = (1/2)*C*V^2 = (1/2)*426*10^-12*(1800)^2 = 691*10^-6 J
Teflon:
The voltage is:
V = 60*10^6* 600*10^-6 = 36*10^3 = 36 kV
According to the listed reference, the relative dielectric constant for teflon is 2.1, this figure multiplies the "ε" of free space.
The capacitance is:
C = ε*A/d = 2.1*8.854*10^-12 * 28.9*10^-3/600*10^-6 = 896*10^-12 F = 896 pF
It would have been easier to note that the capacitance is 2.1 times the air-dielectric case.
The maximum energy stored is:
E = (1/2)*C*V^2 = (1/2)* 896*10^-12* (36*10^3)^2 = 580*10^-3 J
ligma nuts is it the answer bro ;)
Im sure the answer is letter B
