Answer:
Option (e)
Explanation:
A = 45 cm^2 = 0.0045 m^2, d = 0.080 mm = 0.080 x 10^-3 m,
Energy density = 100 J/m
Let Q be the charge on the plates.
Energy density = 1/2 x ε0 x E^2
100 = 0.5 x 8.854 x 10^-12 x E^2
E = 4.75 x 10^6 V/m
V = E x d
V = 4.75 x 10^6 x 0.080 x 10^-3 = 380.22 V
C = ε0 A / d
C = 8.854 x 10^-12 x 45 x 10^-4 / (0.080 x 10^-3) = 4.98 x 10^-10 F
Q = C x V = 4.98 x 10^-10 x 380.22 = 1.9 x 10^-7 C
Q = 190 nC
Vi = As * h = 1000 * 30 = 30,000 cm^3 = Vol. of the ice.
Vb = (Di/Dw) * Vi = (0.9/1.0) * 30,000 = 27,000 cm^3 = Vol. below surface - Vol. of water displaced.
27,000cm^3 * 1g/cm^3 = 27,000 grams = 27 kg = Mass of water displaced.
This question is personal preference... just answer with whatever YOU think, there probably isn’t s wrong answer if I’d guess.
Answer:
Change in Q = 2.1x 10^-3 C
Explanation:
We are given that
The Initialcapacitance C1 = 6.0μF
Initial charge oncapacitor
Q1 = C1 V
= 6.00 x 10^-6 x 100
= 6.00 x 10^-4 C
So the Final capacitance C2 will be
= K x C1 = 4.5 x 6.00 x 10^-6
= 2.7 x 10^ -5 F
So to get Finalcharge
We use Q2 = C2 x V
= 2.7 x 10^ - 5 x 100
= 27 x 10^ -4 C
So Charge flown in thecapacitor is change in Q
Which is = Q2 - Q1
= 27 x 10^-4 - 6.0 x 10^ -4
Change in Q = 2.1x 10^-3 C
Answer:
Explanation:
Effort x effort distance = load x resistance distance
effort distance = 8 m ,
load = 2000N
resistance distance = 1/2 m = 0.5 m
Putting the values in the equation above
effort x 8m = 2000N x .5
effort = 2000 x 0.5 / 8
= 125 N
force required = 125 N .
