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3 years ago

Is a process that modifies light waves so they vibrate in a single plane

Physics

1 answer:

madam [21]3 years ago

8 0

The process you're fishing for is "polarization", but that's a

misleading description.

Polarization doesn't do anything to change the light waves.

It simply filters out (absorbs, as with a polarizing filter) the

light waves that aren't vibrating in the desired plane, and

allows only those that are to pass.

The intensity of a light beam is always reduced after

polarizing it, because much (most) of the original light

has been removed.

A laser light source may be thought of as an exception,

since everything coming out of the laser is polarized.

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Sara is riding her bike down the road at 35 km/hr. She starts to peddle faster as she

steposvetlana [31]

Answer:

Distance covered to top of the hill was : 1.755 km

Explanation:

Initial velocity = 35 km/hr

Acceleration = 2.0 km/hr²

Time taken to accelerate = 3 minutes = 3/60 hours = 1/20 hours

Formula for acceleration : a = Δv /t

v-u/t ---where u is initial velocity , v is final velocity and t is time taken for acceleration

v- 35 / 0.05 = 2

v = 35.10 km/h

Formula for distance is product of speed and time

Distance covered = 35.10 * 0.05 = 1.755 km

6 0

3 years ago

Traveler A starts from rest at a constant acceleration of 6 m/s^2. Two seconds later, traveler B starts with an initial velocity

Troyanec [42]

Answer:

3. 3.5 s

Explanation:

The position of traveller A is given by the equation:

$x_A(t) = \frac{1}{2}a t^2$

where

$a = 6 m/s^2$ is the acceleration of A

t is the time measured from when A started the motion

The position of traveller B instead is given by

$x_B(t) = u_B (t-2) + \frac{1}{2}a(t-2)^2$

where a (acceleration) is the same as traveller A, and

$u_B = 20 m/s$

is B's initial velocity. We can verify that the formula is correct by substituting t=2, and we get $x_B=0$ , which means that B starts its motion 2 seconds later.

Traveller B overtakes traveller A when the two positions are the same, so:

$x_A = x_B\\\frac{1}{2}at^2 = u_B (t-2) + \frac{1}{2}a(t-2)^2\\\frac{1}{2}at^2 = u_B t - 2u_B +\frac{1}{2}at^2 +2a-2at\\u_Bt-2at = 2u_B-2a\\t=\frac{2u_B-2a}{u_B-2a}=\frac{2(20)-2(6)}{20-2(6)}=3.5 s$