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2 years ago

What factor affects the polarization of light by scattering?

Physics

1 answer:

MatroZZZ [7]2 years ago

4 0

Answer:

The correct answer would be A.

Explanation:

I have gotten this answer many times and have never failed once just trust me.

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Which statement correctly explains scientific theories?

Elan Coil [88]

<span>Scientific theories are tested and proven over time; they are then considered scientific laws.

Sometimes however, they are proven wrong, and so they do not become laws

hope this helps</span>

4 0

3 years ago

Read 2 more answers

A box has a 20 N force applied to it to move it 5 m. What is the work done on the box? 4 J 4 N 25 J 100 J

Sunny_sXe [5.5K]

100 J

Explanation:
multiply the force by the distance
20 N x 5 meters = 100 J
please mark brainliest

7 0

3 years ago

Read 2 more answers

ILL GIVE BRAINLIEST!! URGENT PLEASE HELP

anygoal [31]

I believe the answer is the second one.

6 0

3 years ago

Water (density = 1x10^3 kg/m^3) flows at 15.5 m/s through a pipe with radius 0.040 m. The pipe goes up to the second floor of th

RUDIKE [14]

Answer:

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

Explanation:

By assuming that fluid is incompressible and there are no heat and work interaction through the line of current corresponding to the pipe, we can calculate the speed of the water floor in the pipe on the second floor by Bernoulli's Principle, whose model is:

$P_{1} + \frac{\rho\cdot v_{1}^{2}}{2}+\rho\cdot g\cdot z_{1} = P_{2} + \frac{\rho\cdot v_{2}^{2}}{2}+\rho\cdot g\cdot z_{2}$ (1)

Where:

$P_{1}$ , $P_{2}$ - Pressures of the water on the first and second floors, measured in pascals.

$\rho$ - Density of water, measured in kilograms per cubic meter.

$v_{1}$ , $v_{2}$ - Speed of the water on the first and second floors, measured in meters per second.

$z_{1}$ , $z_{2}$ - Heights of the water on the first and second floors, measured in meters.

Now we clear the final speed of the water flow:

$\frac{\rho\cdot v_{2}^{2}}{2} = P_{1}-P_{2}+\rho \cdot \left[\frac{v_{1}^{2}}{2}+g\cdot (z_{1}-z_{2}) \right]$

$\rho\cdot v_{2}^{2} = 2\cdot (P_{1}-P_{2})+\rho\cdot [v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})]$

$v_{2}^{2}= \frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2})$

$v_{2} = \sqrt{\frac{2\cdot (P_{1}-P_{2})}{\rho}+v_{1}^{2}+2\cdot g\cdot (z_{1}-z_{2}) }$ (2)

If we know that $P_{1}-P_{2} = 0\,Pa$ , $\rho=1000\,\frac{kg}{m^{3}}$ , $v_{1} = 15.5\,\frac{m}{s}$ , $g = 9.807\,\frac{m}{s^{2}}$ and $z_{1}-z_{2} = -3.5\,m$ , then the speed of the water flow in the pipe on the second floor is:

$v_{2}=\sqrt{\left(15.5\,\frac{m}{s} \right)^{2}+2\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (-3.5\,m)}$

$v_{2} \approx 13.100\,\frac{m}{s}$

The speed of the water flow in the pipe on the second floor is approximately 13.1 meters per second.

4 0

3 years ago

A softball is thrown from the origin of an X-Y coordinate system with an initial speed of 18 m/s at an angle of 35 degrees above

Goshia [24]

Vo = 18 m/s
angle 35 degrees

1) Components of the initial velocity

Vox = Vo*cos(35) = 18*cos(35) m/s = 14.74 m/s
Voy = Vo* sin(35) = 18*sin(35) m/s = 10.32 m/s

2) Equations of postion:

x = Vox*t
y = Voy*t - gt^2 / 2

3) Calculations

A) t = 0.5 s, t = 1.0 st = 1.5 s, t = 2.0 s

x = 14.74 * t

t = 0.5 s => x = 14.74 m/s * 0.5s = 7.37 m

t = 1.0 s => x = 14.74 m/s * 1.0s = 14.74 m

t = 1.5s => x = 22.11 m

t = 2s => x = 29.48 m

B)

y = Voy*t - gt^2 / 2

Voy = 10.32 m/s
g = 10 m/s (approximation)

y = 10.32*t - 5t^2

t = 0.5 s=> y = 3.91m

t = 1 s => y = 5.32m

t = 1.5 s => y = 4.23m

t = 2 s => y = 0.64 m

7 0

3 years ago