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)
Most of the momentum is transferred to the ball on top. Since the collision in this situation is elastic, momentum is conserved, meaning the momentum of both balls before hitting the floor is equal to the momentum of both balls right after the collision.
Answer:
0.012 J
Explanation:
We are given:
q = 0.0080C
Potential difference = 1.5V
W=qV
Substituting the values into the equation:
W=0.0080*1.5= 0.012J
Y = +1-3 = -2
X = -5+7 = +2
D = √2^2-2^2 = 2√1+1 = 2√2 km
Explanation:
Given that,
Intensity = 1150 W/m²
(a). We need to calculate the magnetic field
Using formula of intensity
![I=\dfrac{E^2}{2\mu_{0}c}](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7BE%5E2%7D%7B2%5Cmu_%7B0%7Dc%7D)
![E=\sqrt{2\times I\mu_{0}c}](https://tex.z-dn.net/?f=E%3D%5Csqrt%7B2%5Ctimes%20I%5Cmu_%7B0%7Dc%7D)
Put the value into the formula
![E=\sqrt{2\times1150\times4\pi\times10^{-7}\times3\times10^{8}}](https://tex.z-dn.net/?f=E%3D%5Csqrt%7B2%5Ctimes1150%5Ctimes4%5Cpi%5Ctimes10%5E%7B-7%7D%5Ctimes3%5Ctimes10%5E%7B8%7D%7D)
![E=931.17\ N/C](https://tex.z-dn.net/?f=E%3D931.17%5C%20N%2FC)
Using formula of magnetic field
![B = \dfrac{E}{c}](https://tex.z-dn.net/?f=B%20%3D%20%5Cdfrac%7BE%7D%7Bc%7D)
Put the value into the formula
![B=\dfrac{931.17}{3\times10^{8}}](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B931.17%7D%7B3%5Ctimes10%5E%7B8%7D%7D)
![B=0.0000031039\ T](https://tex.z-dn.net/?f=B%3D0.0000031039%5C%20T)
![B=3.10\times10^{-6}\ T](https://tex.z-dn.net/?f=B%3D3.10%5Ctimes10%5E%7B-6%7D%5C%20T)
(b). The relative strength of the gravitational and solar electromagnetic pressure forces of the sun on the earth
We need to calculate the gravitational force
Using gravitational force
![F=\dfrac{Gm_{s}M_{e}}{r^2}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7BGm_%7Bs%7DM_%7Be%7D%7D%7Br%5E2%7D)
Put the value into the formula
![F=\dfrac{6.67\times10^{-11}\times1.98\times10^{30}\times5.97\times10^{24}}{(1.496\times10^{11})^2}](https://tex.z-dn.net/?f=F%3D%5Cdfrac%7B6.67%5Ctimes10%5E%7B-11%7D%5Ctimes1.98%5Ctimes10%5E%7B30%7D%5Ctimes5.97%5Ctimes10%5E%7B24%7D%7D%7B%281.496%5Ctimes10%5E%7B11%7D%29%5E2%7D)
![F=3.522\times10^{22}\ N](https://tex.z-dn.net/?f=F%3D3.522%5Ctimes10%5E%7B22%7D%5C%20N)
We need to calculate the radiation force
Using formula of force
![F_{R}=\dfrac{I}{c}\pi\timesR_{E}^{2}](https://tex.z-dn.net/?f=F_%7BR%7D%3D%5Cdfrac%7BI%7D%7Bc%7D%5Cpi%5CtimesR_%7BE%7D%5E%7B2%7D)
Put the value into the formula
![F_{R}=\dfrac{1150}{3\times10^{8}}\times\pi\times(6.378\times10^{6})^2](https://tex.z-dn.net/?f=F_%7BR%7D%3D%5Cdfrac%7B1150%7D%7B3%5Ctimes10%5E%7B8%7D%7D%5Ctimes%5Cpi%5Ctimes%286.378%5Ctimes10%5E%7B6%7D%29%5E2)
![F_{R}=4.8\times10^{8}\ N](https://tex.z-dn.net/?f=F_%7BR%7D%3D4.8%5Ctimes10%5E%7B8%7D%5C%20N)
The gravitational and solar electromagnetic pressure forces of the sun on the earth
![\dfrac{F_{G}}{F_{R}}=\dfrac{3.522\times10^{22}}{4.8\times10^{8}}](https://tex.z-dn.net/?f=%5Cdfrac%7BF_%7BG%7D%7D%7BF_%7BR%7D%7D%3D%5Cdfrac%7B3.522%5Ctimes10%5E%7B22%7D%7D%7B4.8%5Ctimes10%5E%7B8%7D%7D)
![\dfrac{F_{G}}{F_{R}}=7.3375\times10^{13}](https://tex.z-dn.net/?f=%5Cdfrac%7BF_%7BG%7D%7D%7BF_%7BR%7D%7D%3D7.3375%5Ctimes10%5E%7B13%7D)
Hence, This is the required solution.