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Aneli [31]
4 years ago
15

A process that modifies light waves so they viberate in a single plane is refered to as

Physics
1 answer:
PtichkaEL [24]4 years ago
5 0
The question is fishing for "polarization", but the description is all wrong.

Polarization doesn't do anything to the light waves.
It simply blocks any light waves that are not vibrating in
the desired plane, and lets the leftovers pass through.

If there are no light waves already vibrating in the desired plane
in the light that enters the polarizer, then NO LIGHT comes out of it.

If you line up two polarizers and rotate them so that their desired
planes are crossed, then regardless of what goes into the first one,
NO light comes out of the second one !
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A spring gun consists of a spring inside a plastic tube with spring constant, k. The spring can be compressed 20 cm from its equ
emmainna [20.7K]

Answer: The spring constant is K=392.4N/m

Explanation:

According to hook's law the applied force F will be directly proportional to the extension e produced provided the spring is not distorted

The force F=ke

Where k=spring constant

e= Extention produced

h=2m

Given that

e=20cm to meter 20/100= 0.2m

m=100g to kg m=100/1000= 0.1kg

But F=mg

Ignoring air resistance

assuming g=9.81m/s²

Since the compression causes the plastic ball to poses potential energy hence energy stored in the spring

E=1/2ke²=mgh

Substituting our values to find k

First we make k subject of formula

k=2mgh/e²

k=2*0.1*9.81*2/0.1²

K=3.921/0.01

K=392.4N/m

5 0
3 years ago
If the net external force acting on a system is zero , then the total momentum of the system is zero
Neporo4naja [7]
If net external force acting on the system is zero, momentum is conserved. That means, initial and final momentum are same → total momentum of the system is zero.
8 0
4 years ago
Read 2 more answers
Help not to sure with this one need help plz asap​
DiKsa [7]

A is the answer for the problem

4 0
3 years ago
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Determine the velocity required for a moving object 2.00 x 10^4 m above the surface of Mars to escape from Mars's gravity. The m
Ket [755]

Answer:

Therefore the escape velocity from Mar's gravity is 15.88 \times 10^4 m/s.

Explanation:

Escape velocity: Escape velocity is a the minimum velocity that a object needs to escape from the gravitational field of massive body.

V_{escape}=\sqrt{\frac{2GM}{R}}

V_{escape}= Escape velocity

G=Universal gravitational constant = 6.673×10⁻¹¹N m²/Kg²

M= mass of Mars = 6.42×10²³ kg

R = Radius of the Mars = 3.40×10³m

The escape velocity does not depend on the velocity of a object.

V_{escape}=\sqrt{\frac{2\times6.673\times 10^{-11}\times 6.42\times 10^{23}}{3.40\times10^3}}

           =15.88 \times 10^4 m/s

Therefore the escape velocity from Mar's gravity is 15.88 \times 10^4 m/s.

           

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3 years ago
How was your "therapy" session with Eliza? Describe your experience with the therapy program.
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