Answer:
Charge density on the sphere = 2.2 × 10⁻⁸ C/m²
Explanation:
Given:
Radius of sphere (r) = 12 cm = 0.12 m
Distance from the electric field R = 24 cm = 0.24 m
Magnitude (E) = 640 N/C
Find:
Charge density on the sphere
Computation:
Charge on the sphere (q) = (1/K)ER² (K = 9 × 10⁹)
Charge on the sphere (q) = [1/(9 × 10⁹)](640)(0.24)²
Charge on the sphere (q) = 4 × 10⁻⁹ C
Charge density on the sphere = q / [4πr²]
Charge density on the sphere = [4 × 10⁻⁹] / [4(3.14)(0.12)²]
Charge density on the sphere = [4 × 10⁻⁹] / [0.18]
Charge density on the sphere = 2.2 × 10⁻⁸ C/m²
Answer:
A) The event horizon, singularity, and the chute located between the two.
Answer:B,C,D
Explanation:
Thermodynamic efficiency is given by
efficiency can be increased by Keeping 
constant and increasing 
Keeping constant and decreasing 
by increasing 
by decreasing 
ratio
Charge of electron = 1.6×10−¹⁹
(1.6×10−¹⁹)(1×10²) (2e)
= 3.2×10−¹⁷ J
Answer:
The gravitational acceleration of a planet of mass M and radius R
a = G*M/R^2.
In this case we have:
G = 6.67 x 10^-11 N (m/kg)^2
R = 2.32 x 10^7 m
M = 6.35 x 10^30 kg
Now we can compute:
a = (6.67*6.35/2.32^2)x10^(-11 + 30 - 2*7) m/s^2 = 786,907.32 m/s^2
The acceleration does not depend on the mass of the object.
