A mirror is opaque you can not see through it but you can see a reflection within it
Ith air resistance acting on an object that has been dropped, the object will eventually reach a terminal velocity, which is around 53 m/s (195 km/h or 122 mph) for a human skydiver. ... (On the Moon, the gravitational acceleration is much less than on Earth, approximately 1.6 m/s2.)
Again I think you did not give the right constants. So I would use the correct constants for mass of moon and distance from earth to moon.
<span>The formula for force of attraction between any two bodies in the universe
F = GMm / r^2. (Newton's Universal law of Gravitation).
G = Universal gravitational constant, G = 6.67 * 10 ^ -11 Nm^2 / kg^2.
M = Mass of Earth. = 5.97 x 10^24 kg.
m = mass of moon = 7.34 x 10^22 kg.
r = distance apart, between centers = in this case it is the distance from Earth to the Moon
= 3.8 x 10^8 m.
(Sorry I could not assume with the values you gave, they are wrong, and if we use them we would be insulting Physics).
So F = ((6.67 * 10 ^ -11)*(5.97 x 10^24)*(7.34 * 10^22)) / (3.8 x 10^8)^2.
Punch it all up in your calculator.
I used a Casio 991 calculator, it should be one of the best in the world.Really lovely calculator, that has helped me a lot in computations like this. I am thankful for the Calculator.
F = 2.0240 * 10^ 20 N.
So that's our answer.
Hurray!!</span>
Answer:
(a): The magnitude of the electric force on the small sphere =
(b): Shown below.
Explanation:
<u>Given:</u>
- m = mass of the small sphere.
- q = charge on the small sphere.
- L = length of the silk fiber.
- = surface charge density of the large vertical insulating sheet.
<h2>
(a):</h2>
When the dimensions of the sheet is much larger than the distance between the charge and the sheet, then, according to Gauss' law of electrostatics, the electric field experienced by the particle due to the sheet is given as:
<em>where,</em>
is the electrical permittivity of the free space.
The electric field at a point is defined as the amount of electric force experienced by a unit positive test charge, placed at that point. The magnitude electric field at a point and the magnitude of the electric force on a charge q placed at that point are related as:
Thus, the magnitude of the electric force on the small sphere is given by
The sheet and the small sphere both are positively charged, therefore, the electric force between these two is repulsive, which means, the direction of the electric force on the sphere is away from the sheet along the line which is perepndicular to the sheet and joining the sphere.
<h2>
(b):</h2>
When the sphere is in equilibrium, the tension in the fiber is given by the resultant of the weight of the sphere and the electric force experienced by it as shown in the figure attached below.
According to the fig.,
<em>where,</em>
- = electric force on the sphere, acting along left.
- = weight of the sphere, acting vertically downwards.
<em />
g is the acceleration due to gravity.