Answer:
165.8 V/m
Explanation:
The capacitance of a long concentric cylindrical shell of length, L and inner radius, a and outer radius, b is C = 2πε₀L/㏑(b/a)
Since the charge on the cylindrical shells, Q = CV where V = the potential difference across the capacitor(which is the potential difference between the concentric cylindrical shells)
V = Q/C
V = Q ÷ 2πε₀L/㏑(b/a)
V = Q㏑(b/a)/2πε₀L
So, the potential difference per unit length V' is
V' = V/L = Q㏑(b/a)/2πε₀
Given that a = inner radius = 1.5 cm, b = outer radius = 5.6 cm and Q = 7.0 nC = 7.0 × 10⁻⁹ C and ε₀ = 8.854 × 10⁻¹² F/m substituting the values of the variables into the equation, we have
V' = Q㏑(b/a)/2πε₀
V' = 7.0 × 10⁻⁹ C㏑(5.6 cm/1.5 cm)/(2π × 8.854 × 10⁻¹² F/m)
V' = 7.0 × 10⁻⁹ C㏑(3.733)/(55.631 × 10⁻¹² F/m)
V' = 7.0 × 10⁻⁹ C × 1.3173/(55.631 × 10⁻¹² F/m)
V' = 9.2211 × 10⁻⁹ C/(55.631 × 10⁻¹² F/m)
V' = 0.16575 × 10³ V/m
V' = 165.75 V/m
V' ≅ 165.8 V/m
Answer:31.3W
Explanation: JUST A FORMULA
P = V*V/R = 120*120/460 = 31.3W
<h3> it depends on numbers of turns coils area restoring force per unit twist and magnetic fields</h3><h2 /><h2 /><h2>hope it helps</h2>
Answer:
The speed of the projectile upon returning to its starting point is 50 m/s.
Explanation:
Given;
initial velocity of the projectile, u = 50 m/s
The velocity of the projectile is maximum before hitting the ground. As the object moves upward, the velocity reduces as it approaches the maximum height, at the maximum height the velocity becomes zero. Also, as the object moves downward, the velocity starts to increase and becomes maximum before hitting the ground.
Therefore, the speed of the projectile upon returning to its starting point is 50 m/s.
The answer is 7.93 pounds
