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In-s [12.5K]
3 years ago
6

Originally, light is unpolarized. It then travels through 3 polarizers: the first with its transmission axis aligned vertical, t

he next at a 30 degree angle to the vertical and the last aligned at a 45 degree angle to the vertical (tilted the same direction as the first with respect to the vertical). The original intensity of light is 4 W/m2. 3) What is the intensity of light after the first polarizer? A. 4 W/m2 B. 2 W/m2 C. 1.5 W/m2 D. 1.4 W/m2 E. 1.0 W/m2 F. 0 W/m2
Physics
1 answer:
Ilya [14]3 years ago
5 0

Answer:

C. 1.5 W/m²

Explanation:

I_0 = Unpolarized light = 4 W/m²

\theta = Angle with which the polarizer is turned

The polarization of light is given by

I=\frac{I_0}{2}cos^2\theta\\\Rightarrow I=\frac{4}{2}cos^2(30)\\\Rightarrow I=2\times \left(\frac{\sqrt3}{2}\right)^2\\\Rightarrow I=1.5\ W/m^2

The intensity of light after passing through the first polarizer is 1.5 W/m²

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When a 0.622 kg basketball hits the floor, its velocity changes from 4.23 m/s down to 3.85 m/s up. If the ball was in contact wi
SVEN [57.7K]

when the ball hits the floor and bounces back the momentum of the ball changes.

the rate of change of momentum is the force exerted by the floor on it.

the equation for the force exerted is

f = rate of change of momentum

f = \frac{mv - mu}{t}

v is the final velocity which is - 3.85 m/s

u is initial velocity - 4.23 m/s

m = 0.622 kg

time is the impact time of the ball in contact with the floor - 0.0266 s

substituting the values

f = \frac{0.622 kg (3.85 m/s - (-)4.23 m/s)}{0.0266}

since the ball is going down, we take that as negative and ball going upwards as positive.

f = 189 N

the force exerted from the floor is 189 N

4 0
3 years ago
O'Malley is riding on a bus which is moving at 10 m/s, and he throws a ball which he observes to be moving at 10 m/s relative to
Vikki [24]

Answer:

<em>20 m/s in the same direction of the bus.</em>

Explanation:

<u>Relative Motion </u>

Objects movement is always related to some reference. If you are moving at a constant speed, all the objects moving with you seem to be at rest from your reference, but they are moving at the same speed as you by an external observer.

If we are riding on a bus at 10 m/s and throw a ball which we see moving at 10 m/s in our same direction, then an external observer (called Ophelia) will see the ball moving at our speed plus the relative speed with respect to us, that is, at 20 m/s in the same direction of the bus.

3 0
3 years ago
What happens to the temperature during an endothermic reaction
Aleksandr-060686 [28]

Exothermich cools and endo heats so it heats

7 0
3 years ago
Read 2 more answers
Suppose the gas resulting from the sublimation of 1.00 g carbon dioxide is collected over water at 25.0◦c into a 1.00 l containe
AlexFokin [52]

Answer:

0.56 atm

Explanation:

First of all, we need to find the number of moles of the gas.

We know that

m = 1.00 g is the mass of the gas

Mm=44.0 g/mol is the molar mass of the carbon dioxide

So, the number of moles of the gas is

n=\frac{m}{M_m}=\frac{1.00 g}{44.0 g/mol}=0.023 mol

Now we can find the pressure of the gas by using the ideal gas equation:

pV=nRT

where

p is the pressure

V=1.00 L = 0.001 m^3 is the volume

n = 0.023 mol is the number of moles

R=8.314 J/mol K is the gas constant

T=25.0^{\circ}+273=298 K is the temperature of the gas

Solving the equation for p, we find

p=\frac{nRT}{V}=\frac{(0.023 mol)(8.314 J/mol K)(298 K)}{0.001 m^3}=5.7 \cdot 10^4 Pa

And since we have

1 atm = 1.01\cdot 10^5 Pa

the pressure in atmospheres is

p=\frac{5.7\cdot 10^4 Pa}{1.01\cdot 10^5 Pa/atm}=0.56 atm

5 0
4 years ago
Water is flowing in a pipe with a circular cross section but with varying cross-sectional area, and at all points the water comp
slamgirl [31]

(a) 5.66 m/s

The flow rate of the water in the pipe is given by

Q=Av

where

Q is the flow rate

A is the cross-sectional area of the pipe

v is the speed of the water

Here we have

Q=1.20 m^3/s

the radius of the pipe is

r = 0.260 m

So the cross-sectional area is

A=\pi r^2 = \pi (0.260 m)^2=0.212 m^2

So we can re-arrange the equation to find the speed of the water:

v=\frac{Q}{A}=\frac{1.20 m^3/s}{0.212 m^2}=5.66 m/s

(b) 0.326 m

The flow rate along the pipe is conserved, so we can write:

Q_1 = Q_2\\A_1 v_1 = A_2 v_2

where we have

A_1 = 0.212 m^2\\v_1 = 5.66 m/s\\v_2 = 3.60 m/s

and where A_2 is the cross-sectional area of the pipe at the second point.

Solving for A2,

A_2 = \frac{A_1 v_1}{v_2}=\frac{(0.212 m^2)(5.66 m/s)}{3.60 m/s}=0.333 m^2

And finally we can find the radius of the pipe at that point:

A_2 = \pi r_2^2\\r_2 = \sqrt{\frac{A_2}{\pi}}=\sqrt{\frac{0.333 m^2}{\pi}}=0.326 m

6 0
3 years ago
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