the force applied when using a simple machine is called the effort force
Answer:

Explanation:
When the unpolarized light passes through the first polarizer, only the component of the light parallel to the axis of the polarizer passes through.
Therefore, after the first polarizer, the intensity of light passing through it is halved, so the intensity after the first polarizer is:

Then, the light passes through the second polarizer. In this case, the intensity of the light passing through the 2nd polarizer is given by Malus' law:

where
is the angle between the axes of the two polarizer
Here we have

So the intensity after the 2nd polarizer is

And substituting the expression for I1, we find:

Answer:
about 4.74 seconds
Explanation:
The time to fall distance d from height h is given by ...
t = √(2d/g)
t = √(2·110 m/(9.8 m/s^2)) ≈ 4.74 s
It will take the car about 4.74 seconds to fall 110 meters to the river.
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We assume the car's speed is horizontal, so does not add or subtract anything to/from the time to fall from the height.
The Kepler's laws predict the planetary motion, so there are three laws for this, namely:
1. The orbit of a planet is an ellipse with the Sun (the sun is a star!) at one of the two focus.
2. A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
3. The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.
So, let's use second law. The Sun sweeps out equal areas during equal intervals of time means that if A = B, the time the planet takes to travel A1A2 is equal to the time the planet takes to travel B1B2, but given that A = 2B, then takes twice the time to travel A1A2 compared to B1B2.
Answer:
The new distance is d = 0.447 d₀
Explanation:
The electric out is given by Coulomb's Law
F = k q₁ q₂ / r²
This electric force is in balance with tension.
We reduce the charge of sphere B to 1/5 of its initial value (
=q₂ = q₂ / 5) than new distance (d = n d₀)
dat
q₁ = 
q₂ = 
r = d₀
In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points
F = k q₁ q₂ / d₀²
F = k q₁ (q₂ / 5) / (n d₀)²
.k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)
5 n² = 1
n = √ 1/5
n = 0.447
The new distance is
d = 0.447 d₀