Answer: distance d = 4.73e10m
Explanation: Suppose the charge on the black hole is 5740 C which is a positive charge.
Using electric potential V formula:
V = kq / d
Where K = 9.05×10^9Nm^2/C
And e = 1.6×10^-19C
But you don't need to substitute it.
1090 V = 8.99e9N·m²/C² * 5740C /d
Make d the subject of formula
d = 4.73e10 m
In some early mornings , dew drops can be found on grass or a car parked outside, but not on other materials such as the sidewalk because the night -time temperature on grass and the car went below the dew point, but the temperature of the concrete did not drop enough to reach the dew point level
Dew can be formed on any object when the temperature of the object drop. When this happen, the object will be cool which will eventually cool the surrounding air around the object.
Dew drops is as a result of condensation in the air. When the cool air causes the air vapor to convert to liquid. The dew will form when the temperature of the object balances with the dew point in the surrounding environment.
In some early mornings , dew drops can be found on grass or a car parked outside, but not on other materials such as the sidewalk because the night -time temperature on grass and the car went below the dew point, but the temperature of the concrete did not drop enough to reach the dew point level
Therefore the correct option is therefore A
Learn more here : brainly.com/question/13834972
The strong nuclear force overcomes the electric force of repulsion thatacts among the protons in thenucleus. B. The weak nuclear force is involved in certain types of radioactive processes. A.The strong nuclear force is a powerful force of attraction that acts only on theneutrons and protons in the nucleus.
Answer:
(a) q = 2.357 x 10⁻⁵ C
(b) Φ = 2.66 x 10⁶ N.m²/C
Explanation:
Given;
diameter of the sphere, d = 1.1 m
radius of the sphere, r = 1.1 / 2 = 0.55 m
surface charge density, σ = 6.2 µC/m²
(a) Net charge on the sphere
q = 4πr²σ
where;
4πr² is surface area of the sphere
q is the net charge on the sphere
σ is the surface charge density
q = 4π(0.55)²(6.2 x 10⁻⁶)
q = 2.357 x 10⁻⁵ C
(b) the total electric flux leaving the surface of the sphere
Φ = q / ε
where;
Φ is the total electric flux leaving the surface of the sphere
ε is the permittivity of free space
Φ = (2.357 x 10⁻⁵) / (8.85 x 10⁻¹²)
Φ = 2.66 x 10⁶ N.m²/C
