They only conduct when they are in solution form, because then their ions become mobile, and the ions conduct electricity.
Hope this helps!
Answer: The unpolarized light's intensity is reduced by the factor of two when it passes through the polaroid and becomes linearly polarized in the plane of the Polaroid. When the polarized light passes through the polaroid with the plane of polarization at an angle
with respect to the polarization plane of the incoming light, the light's intensity is reduced by the factor of
(this is the Law of Malus).
Explanation: Let us say we have a beam of unpolarized light of intensity
that passes through two parallel Polaroid discs with the angle of
between their planes of polarization. We are asked to find
such that the intensity of the outgoing beam is
. To solve this we follow the steps below:
Step 1. It is known that when the unpolarized light passes through a polaroid its intensity is reduced by the factor of two, meaning that the intensity of the beam passing through the first polaroid is

This beam also becomes polarized in the plane of the first polaroid.
Step 2. Now the polarized beam hits the surface of the second polaroid whose polarization plane is at an angle
with respect to the plane of the polarization of the beam. After passing through the polaroid, the beam remains polarized but in the plane of the second polaroid and its intensity is reduced, according to the Law of Malus, by the factor of
This yields
. Substituting from the previous step we get

yielding

and finally,

The spring starts out 22 cm long with nothing hanging on it.
Hanging 35 newtons of weight on it stretches the spring 1 meter.
Ellen is going to hang 250 grams of mass on the spring.
What's the weight of 250 grams of mass ?
Weight = (mass) x (acceleration of gravity in the place where the mass is) .
On Earth, the acceleration of gravity is 9.8 m/s² .
250 grams is 0.25 of a kilogram.
Weight of 250 grams = (0.25 kilogram) x (9.8 m/s²)
= (0.25 x 9.8) kg-m/s²
= 2.45 newtons .
2.45 newtons of weight is (2.45 / 35) of 35 newtons,
so it'll stretch the spring (2.45 / 35) of a meter.
2.45/35 = 0.07 of a meter = 7 centimeters.
The spring was 22 cm long with nothing hanging on it,
and the 250-gm weight stretched it 7 cm.
So with the weight hanging on it, it's (22 + 7) = 29 cm long.
Answer:
3. 0.5 sec.
Explanation:
A bullet fired horizontally follows a projectile motion, which consists of two independent motions:
- A horizontal motion with constant speed
- A vertical motion with constant acceleration, g = 9.8 m/s^2, towards the ground
The time taken for the bullet to reach the ground can be calculated just by considering the vertical motion:

where y is the vertical position at time t, h is the initial height, and
is the initial vertical velocity of the bullet.
Since the bullet is fired horizontally,
. So the equation becomes

And the time that the bullet takes to reach the ground can be found by requiring y=0 and solving for t:

As we can see, in this equation there is no dependance on the initial speed of the bullet: therefore, if the bullet is fired still horizontally but with a different speed, it will still take the same time (0.5 s) to reach the ground.