Answer:
A) 1.4167 × 10^(-11) F
B) r_a = 0.031 m
C) E = 3.181 × 10⁴ N/C
Explanation:
We are given;
Charge;Q = 3.40 nC = 3.4 × 10^(-9) C
Potential difference;V = 240 V
Inner radius of outer sphere;r_b = 4.1 cm = 0.041 m
A) The formula for capacitance is given by;
C = Q/V
C = (3.4 × 10^(-9))/240
C = 1.4167 × 10^(-11) F
B) To find the radius of the inner sphere,we will make use of the formula for capacitance of spherical coordinates.
C = (4πε_o)/(1/r_a - 1/r_b)
Rearranging, we have;
(1/r_a - 1/r_b) = (4πε_o)/C
ε_o is a constant with a value of 8.85 × 10^(−12) C²/N.m
Plugging in the relevant values, we have;
(1/r_a - 1/0.041) = (4π × 8.85 × 10^(−12) )/(1.4167 × 10^(-11))
(1/r_a) - 24.3902 = 7.8501
1/r_a = 7.8501 + 24.3902
1/r_a = 32.2403
r_a = 1/32.2403
r_a = 0.031 m
C) Formula for Electric field just outside the surface of the inner sphere is given by;
E = kQ/r_a²
Where k is a constant value of 8.99 × 10^(9) Nm²/C²
Thus;
E = (8.99 × 10^(9) × 3.4 × 10^(-9))/0.031²
E = 3.181 × 10⁴ N/C
Answer:
α = 1.875 rad/s²
t = 8 s
Explanation:
α = ω²/2θ = 15²/(2(60)) = 1.875 rad/s²
t = ω/a = 15 / 1.875 = 8 s
Explanation:
The height of the rise of liquid with capillary tube is given by the formula as follows :
Where
r is radius
It is clear that the height of the rise of liquid is inversely proportional to the radius of the capillary tube.
If the radius of the capillary tube is doubled, it means the height of rise of liquid with capillary tube become half.
I’m sure gravitation should be at the river sorry if I’m wrong
Answer:
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Explanation:
From the exercise we have our initial information
When the balloon gets to the ceiling its velocity at that moment is 0 m/s. Being said that we can calculate velocity at the vertical direction
Since 
and 
Knowing that
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