Answer:

Explanation:
Given that:
- Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
- separation distance of capacitor 2,

- separation distance of capacitor 1,

- quantity of charge on capacitor 2,

- quantity of charge on capacitor 1,

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.
Mathematically given as:
.....................................(1)
where:
k = relative permittivity of the dielectric material between the plates= 1 for air

From eq. (1)
For capacitor 2:

For capacitor 1:

![C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]](https://tex.z-dn.net/?f=C_1%3D%5Cfrac%7B1%7D%7B2%7D%20%5B%20%5Cfrac%7Bk.%5Cepsilon_0.A%7D%7Bd%7D%5D)
We know, potential differences across a capacitor is given by:
..........................................(2)
where, Q = charge on the capacitor plates.
for capacitor 2:


& for capacitor 1:


![V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]](https://tex.z-dn.net/?f=V_1%3D8%5Ctimes%20%5B%5Cfrac%7BQ.d%7D%7Bk.%5Cepsilon_0.A%7D%5D)

Answer:
3560.36 Watts
Explanation:
Power,
where P is power, n is the number of skiers, t is time in seconds and Δt is change in time, ΔW is given by mgh where m is mass, g is gravitational constant, h is height
Substituting n for 4 skiers, m for 62.9 Kg, g for 9.81, h for 148 m and t for 1.71*60=102.6 seconds
P=
Average power is approximately 3560.36 Watts
Answer:
31.905 ft/s²
Explanation:
Given that
Mass of the pilot, m = 120 lb
Weight of the pilot, w = 119 lbf
Acceleration due to gravity, g = 32.05 ft/s²
Local acceleration of gravity of found by using the relation
Weight in lbf = Mass in lb * (local acceleration/32.174 lbft/s²)
119 = 120 * a/32. 174
119 * 32.174 = 120a
a = 3828.706 / 120
a = 31.905 ft/s²
Therefore, the local acceleration due to gravity at that elevation is 31.905 ft/s²
Answer:
a) v, v
b) 2mv^2
c) Elastic collion
Explanation:
(a) The velocity of the second particle after the collision is (v2x,v2y)=(v,−v). From momentum conservation in x-direction
Here x, y represent direction.They are not variable. 1 and 2 represent before and after.
2vm=v1xm+v2xm, we find v1x=v.
From momentum conservation in y-direction
0 =v1ym+v2ym, we findv1y=v.
(b) By energy conservation principle
Before: K=1/2m(2v)^2=2mv^2.
After: K=1/2m(v^2(1x)+v^2(1y))+12m(v22x+v22y)=2mv^2
(c) The collision is elastic
The answer is gravity. I hope this helps.