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tiny-mole [99]
3 years ago
8

Unpolarized light is passed through an optical filter that is oriented in the vertical direction. 1) If the incident intensity o

f the light is 90 W/m2, what is the intensity of the light that emerges from the filter? (Express your answer to two significant figures.)
Physics
1 answer:
lions [1.4K]3 years ago
6 0

Answer:

45 W/m^2

Explanation:

Intensity of light, Io = 90 W/m^2

According to the law of Malus

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

The average value of Cos^θ is half

So, I = Io/2

I = 90 /2

I = 45 W/m^2

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An arrow is launched upward with an initial speed of 100 meters per second (m/s). The equations above describe the constant-acce
ankoles [38]

Answer:

d=510.2m

t=10.2s

Explanation:

The formulas for accelerated motion are:

v=v_0+at\\x=x_0+v_0t+\frac{at^2}{2}

From them we can get v^2=v_0^2+2ad.

We have:

v-v_0=at\\t=\frac{v-v_0}{a}

And substitute:

x=x_0+v_0(\frac{v-v_0}{a})+\frac{a}{2}(\frac{v-v_0}{a})^2\\x-x_0=\frac{v_0(v-v_0)}{a}+\frac{(v-v_0)^2}{2a}

We multiply both sides by 2a, and continue:

2a(x-x_0)=2v_0(v-v_0)+(v-v_0)^2=2v_0v-2v_0^2+v^2+v_0^2-2vv_0=v^2-v_0^2

Being d the displacement x-x_0, we have v^2=v_0^2+2ad

For our exercise, we will write this as:

d=\frac{v^2-v_0^2}{2a}

And taking upwards direction positive and imposing final velocity 0m/s (for maximum height), we have:

d=\frac{-v_0^2}{2a}=\frac{-(100m/s)^2}{2(-9,8m/s^2)}=510.2m

For the time we use:

t=\frac{v-v_0}{a}=\frac{-v_0}{a}=\frac{-(100m/s)}{(-9.8m/s^2)}=10.2s

6 0
3 years ago
Explain whether there can be forces act-
Xelga [282]

Technically friction is acting on the car because it is still rubbing against the street and gravity is pulling the car down preventing it from floating??? lol

5 0
2 years ago
Where would you expect to have more touch receptors: on the palm of your hand or on the back of your hand? Explain your reasonin
anygoal [31]

Answer:

ive answered this

Explanation:

please check your previose question

6 0
3 years ago
A projectile is launched at an angle of 29 degrees above the horizontal with an initial velocity of 36.6 at an unknown height.
alex41 [277]

The magnitude of the unknown height of the projectile is determined as 16.1 m.

<h3>Magnitude of the height</h3>

The magnitude of the height of the projectile is calculated as follows;

H = u²sin²θ/2g

H = (36.6² x (sin 29)²)/(2 x 9.8)

H = 16.1 m

Thus, the magnitude of the unknown height of the projectile is determined as 16.1 m.

Learn more about height here: brainly.com/question/1739912

#SPJ1

3 0
2 years ago
A sealed tank containing seawater to a height of 10.5 mm also contains air above the water at a gauge pressure of 2.95 atmatm. W
weqwewe [10]

Answer:

The water is flowing at the rate of 28.04 m/s.

Explanation:

Given;

Height of sea water, z₁ = 10.5 m

gauge pressure, P_{gauge \ pressure} = 2.95 atm

Atmospheric pressure, P_{atm} = 101325 Pa

To determine the speed of the water, apply Bernoulli's equation;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 = P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2

where;

P₁ = P_{gauge \ pressure} + P_{atm \ pressure}

P₂ = P_{atm}

v₁ = 0

z₂ = 0

Substitute in these values and the Bernoulli's equation will reduce to;

P_1 + \rho gz_1 + \frac{1}{2}\rho v_1^2 =  P_2 + \rho gz_2 + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 + \frac{1}{2}\rho (0)^2 =  P_2 + \rho g(0) + \frac{1}{2}\rho v_2^2\\\\P_1 + \rho gz_1 =  P_2 + \frac{1}{2}\rho v_2^2\\\\P_{gauge} + P_{atm} + \rho gz_1 = P_{atm} + \frac{1}{2}\rho v_2^2\\\\P_{gauge} +  \rho gz_1 =  \frac{1}{2}\rho v_2^2\\\\v_2^2 = \frac{2(P_{gauge} +  \rho gz_1)}{\rho} \\\\v_2 = \sqrt{ \frac{2(P_{gauge} +  \rho gz_1)}{\rho} }

where;

\rho is the density of seawater = 1030 kg/m³

v_2 = \sqrt{ \frac{2(2.95*101325 \ + \  1030*9.8*10.5 )}{1030} }\\\\v_2 = 28.04 \ m/s

Therefore, the water is flowing at the rate of 28.04 m/s.

7 0
3 years ago
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