Answer:
a. 
b.
must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping
c.
is the time taken to stop after braking
Explanation:
Given:
- speed of leading car,

- speed of lagging car,

- distance between the cars,

- deceleration of the leading car after braking,

a.
Time taken by the car to stop:

where:
, final velocity after braking
time taken


b.
using the eq. of motion for the given condition:

where:
final velocity of the chasing car after braking = 0
acceleration of the chasing car after braking

must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping
c.
time taken by the chasing car to stop:


is the time taken to stop after braking
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:
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Explanation:
Answer:
Explanation:
you can seperate mixtures by filtration, evaporation, distillation and chromatography.