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Troyanec [42]
4 years ago
11

What angle is needed between the direction of polarized light and the axis of a polarizing filter to cut its intensity in half?

Physics
1 answer:
cricket20 [7]4 years ago
8 0

Answer:

45 degree

Explanation:

Let the initial intensity is Io and the angle between the polarizer and the axis of filter is θ.

The intensity of light passing after filter is 0.5 Io.

By using the law of Malus,

I=I_{0} Cos^{2}\theta

0.5I_{0}=I_{0} Cos^{2}\theta

0.5= Cos^{2}\theta

0.707= Cos\theta

θ = 45 degree

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The gravitational acceleration on Mars is 3.71 m/s2. If the pendulums were set in motion on the red planet, how would that affec
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On Earth, the period of a pendulum is given by:
T_{earth}=2\pi  \sqrt{ \frac{L}{g_{earth} }
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Similarly, the period of the same pendulum on Mars will be
T_{mars}=2\pi \sqrt{ \frac{L}{g_{mars} }
where g_{mars}=3.71~m/s^2 is the gravitational acceleration on Mars.
Therefore, if we want to see how does the period of the pendulum on Mars change compared to the one on Earth, we can do the ratio between the two of them:
\frac{T_{mars}}{T_{earth}}= \sqrt{ \frac{g_{earth}}{g_{mars}} }  =  \sqrt{ \frac{9.81~m/s^s}{3.71~m/s^2} }=1.63
Therefore, the period of the pendulum on Mars will be 1.63 times the period on Earth.
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4 years ago
Two cars travel on a straight road from point A to point B. Both cars accelerate to their maximum speed and then continue at tha
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B) Car 2 has both a larger acceleration and a larger average speed.

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3 years ago
The mass of a hypothetical planet is 1/100 that of Earth and its radius is 1/4 that of Earth. If a person weighs 600 N on Earth,
omeli [17]

To solve this problem we will apply the Newtonian concept of gravitational acceleration produced by a planet. This relationship is given by:

g = \frac{GM}{r^2}

Where,

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius

The values given are based on the constants of the earth, so they can be expressed as

M_p = \frac{1}{100} M_e

r_p = \frac{1}{4} r_e

The relationship of gravity would then be given:

g_e = \frac{GM_e}{r_e^2}

The relationship with the new planet, from the gravity of the earth would be given

g_p = \frac{GM_p}{r_p^2}

g_p = \frac{G(1/100)M_e}{(1/4 r_e)^2}

g_p = \frac{GM_e 16}{100 r_e^2}

g_p = 0.16 \frac{GM_e}{r_e^2}

g_p = 0.16g_e

The relationship with the weight of the earth would be given as:

W_e = m*g_e = 600N

W_p = m*g_p = m(0.16g_p)

W_p = (m*g_p)(0.16)

W_p = 600*0.16

W_p = 96N

Therefore the weigh on this planet would be 96N

3 0
3 years ago
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