Answer:
The electric field is 
Explanation:
From the question we are told that
The radius of the metal sphere is 
The excess charge which the metal sphere carries is 
The distance of the position being to the center is 
The coulomb constant is 
Generally the electric field is mathematically represented as
substituting values
Answer:
1.129×10⁻⁵ N
1.295 m
Explanation:
Take right to be positive. Sum of forces on the 31.8 kg mass:
∑F = GM₁m / r₁² − GM₂m / r₂²
∑F = G (M₁ − M₂) m / r²
∑F = (6.672×10⁻¹¹ N kg²/m²) (516 kg − 207 kg) (31.8 kg) / (0.482 m / 2)²
∑F = 1.129×10⁻⁵ N
Repeating the same steps, but this time ∑F = 0 and we're solving for r.
∑F = GM₁m / r₁² − GM₂m / r₂²
0 = GM₁m / r₁² − GM₂m / r₂²
GM₁m / r₁² = GM₂m / r₂²
M₁ / r₁² = M₂ / r₂²
516 / r² = 207 / (0.482 − r)²
516 (0.482 − r)² = 207 r²
516 (0.232 − 0.964 r + r²) = 207 r²
119.9 − 497.4 r + 516 r² = 207 r²
119.9 − 497.4 r + 309 r² = 0
r = 0.295 or 1.315
r can't be greater than 0.482, so r = 0.295 m.
Answer:
I = 0.96 A
Explanation:
No of electrons, 
Time, t = 3 ms = 
We need to find the electric current. We know that electric charge per unit time is equal to the electric current.
q = ne (Quantization of electric charge)
So, the electric current is 0.96 A.
This is a problem of conservation of momentum
Momentum before throwing the rock: m*V = 96.0 kg * 0.480 m/s = 46.08 N*s
A) man throws the rock forward
=>
rock:
m1 = 0.310 kg
V1 = 14.5 m/s, in the same direction of the sled with the man
sled and man:
m2 = 96 kg - 0.310 kg = 95.69 kg
v2 = ?
Conservation of momentum:
momentum before throw = momentum after throw
46.08N*s = 0.310kg*14.5m/s + 95.69kg*v2
=> v2 = [46.08 N*s - 0.310*14.5N*s ] / 95.69 kg = 0.434 m/s
B) man throws the rock backward
this changes the sign of the velocity, v2 = -14.5 m/s
46.08N*s = - 0.310kg*14.5m/s + 95.69kg*v2
v2 = [46.08 N*s + 0.310*14.5 N*s] / 95.69 k = 0.529 m/s
