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Darya [45]
3 years ago
8

Two polarizers are held in front of a light source such that their transmission axes are parallel. As the angle between the two

transmission axes increases toward 90°, which of the following is true?
A.The intensity of the transmitted light increases.

B.The intensity of the transmitted light decreases.

C.The amount of light reflected back to the source increases

.D.The amount of light reflected back to the source decreases.
Physics
1 answer:
nikklg [1K]3 years ago
4 0
I think it’s B, because held with the transmission axes perpendicular to each other,
no light will pass through the films
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Answer:

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Explanation:

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3 years ago
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A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string h
hram777 [196]

Answer:

v_{2}=3.5 m/s

Explanation:

Using the conservation of energy we have:

\frac{1}{2}mv^{2}=mgh

Let's solve it for v:

v=\sqrt{2gh}

So the speed at the lowest point is v=7 m/s

Now, using the conservation of momentum we have:

m_{1}v_{1}=m_{2}v_{2}

v_{2}=\frac{1*7}{2}

Therefore the speed of the block after the collision is v_{2}=3.5 m/s

I hope it helps you!

       

8 0
3 years ago
Who showed that our universe is heliocentric—the planets of the solar system revolve around the sun? Johannes Kepler Isaac Newto
denis23 [38]

Answer:

Option (3)

Explanation:

Nicolaus Copernicus was an astronomer from Poland, who was born on the 19th of February in the year 1473. He played a great role in the field of modern astronomy.

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6 0
3 years ago
A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum
s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle=q=-2.74\times 10^{-6} C

Kinetic energy of particle=K_E=6.65\times 10^{-10} J

Initial time=t_1=6.36 s

Final potential difference=V_2=0.351 V

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

qV=K.E

Using the formula

2.74\times 10^{-6}V_1=6.65\times 10^{-10} J

V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V

Initial voltage=V_1=2.43\times 10^{-4} V

\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2

Using the formula

\frac{V_1}{V_2}=(\frac{6.36}{t})^2

\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}

t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}

t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}

t=241.7 s

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0
3 years ago
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