Answer:
C1/C2 = 0.213 or C2/C1 = 4.68
Explanation:
Please refer to the attached image for step by step explanation.
Answer:
W = 9.93 10² N
Explanation:
To solve this exercise we must use the concept of density
ρ = m / V
the tabulated density of copper is rho = 8966 kg / m³
let's find the volume of the cylindrical tube
V = A L
V = π (R_ext ² - R_int ²) L
let's calculate
V = π (4² - 2²) 10⁻⁴ 3
V = 1.13 10⁻² m³
m = ρ V
m = 8966 1.13 10⁻²
m = 1.01 10² kg
the weight of the tube
W = mg
W = 1.01 10² 9.8
W = 9.93 10² N
T<span>he equation to be used here to determine the distance between two equipotential points is:
V = k * Q / r
where v is the voltage of the point, k is a constant, Q is charge of the point measured in coloumbs and r is the distance.
In this case, we can use ratio of proportions to determine the distance between the two points. in this respect,
Point 1:
V = k * Q / r = 290
r = k*Q/290 ; kQ = 290r
Point 2:
V = k * Q / R = 41
R = k*Q/41
from equation 10 kQ = 290r
R = 290/(41)= 7.07 m
The distance between the two points then is equal to 7.07 m.
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The velocity is 6.75
The velocity in the equation stated above can be calculated as follows
m= 2,000
p= 2.25
y= 6000
velocity= 2.25 × 6000/ 2000
= 13500/2000
= 6.75
Hence the velocity is 6.75
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