Machines makes work easier by increasing the amount of force that is applied, and changing the direction in which the force is applied !! Hope it helped (p.s. I had this same question)
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:
We know, potential differences across a capacitor is given by:
..........................................(2)
where, Q = charge on the capacitor plates.
for capacitor 2:
& for capacitor 1:
Answer: a) 139.4 μV; b) 129.6 μV
Explanation: In order to solve this problem we have to use the Ohm law given by:
V=R*I whre R= ρ *L/A where ρ;L and A are the resistivity, length and cross section of teh wire.
Then we have:
for cooper R=1.71 *10^-8* 1.8/(0.001628)^2= 11.61 * 10^-3Ω
and for silver R= 1.58 *10^-8* 1.8/(0.001628)^2=10.80 * 10^-3Ω
Finalle we calculate the potential difference (V) for both wires:
Vcooper=11.62* 10^-3* 12 * 10^-3=139.410^-6 V
V silver= 10.80 10^-3* 12 * 10^-3=129.6 10^-6 V
Answer:
frequency of the sound = f = 1,030.3 Hz
phase difference = Φ = 229.09°
Explanation:
Step 1: Given data:
Xini = 0.540m
Xfin = 0.870m
v = 340m/s
Step 2: frequency of the sound (f)
f = v / λ
λ = Xfin - Xini = 0.870 - 0.540 = 0.33
f = 340 / 0.33
f = 1,030.3 Hz
Step 3: phase difference
phase difference = Φ
Φ = (2π/λ)*(Xini - λ) = (2π/0.33)* (0.540-0.33) = 19.04*0.21 = 3.9984
Φ = 3.9984 rad * (360°/2π rad)
Φ = 229.09°
Hope this helps!