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alexgriva [62]
3 years ago
7

The sine of the incident angle is 0.217; the sine of the refracted angle is 0.173. Calculate the index of refraction.

Physics
2 answers:
Tju [1.3M]3 years ago
6 0

Answer:

1.3.

Explanation:

From the question given above, the following data were obtained:

Sine of incident angle (Sine i) = 0.217

Sine of refracted angle (Sine r) = 0.173

Refractive index (n) =?

From snell's law, the Refractive index is simply defined as the ratio of the sine of the incident angle to sine of the refracted angle. Mathematically, it is expressed as:

Refractive index = Sine of incident angle / sine of refracted angle

n = Sine i / Sine r

With the above formula, we can obtain index of refraction as shown below:

Sine of incident angle (Sine i) = 0.217

Sine of refracted angle (Sine r) = 0.173

Refractive index (n) =?

n = Sine i / Sine r

n = 0.217 / 0.173

n = 1.3

Thus, the index of refraction is 1.3.

Ksenya-84 [330]3 years ago
6 0

Answer:

n = 1.25

Explanation:

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ludmilkaskok [199]

Answer:

Explanation:

Assuming school is at the end of the 20 mile route, then

20 mi / 35 mi/hr = 0.57142...hr

which is about 34 minutes 17 seconds

6 0
3 years ago
A manufacturer claims its cleanser works twice as fast as any other. Could test be performed to support the claim? Explain
lora16 [44]

Yes, a test could be performed to support the claim.

 

Hypothesis: The claim that a manufacturer’s cleanser works twice as fast as any other cleanser.

 

So, based from this hypothesis, we can perform the following tests:

We assign Cleanser A to the manufacturer that claims that their cleanser works twice as fast as any other cleanser and Cleanser B to the cleanser to be compared with.

 

1.       Get two tiles and put the same amount of stain on them.

2.       Apply Cleanser A on the first tile and Cleanser B on the second tile.

3.       Apply the same amount of force in removing the stains on both tiles

4.       Record the amount of time it takes to remove the stains on each tile.

4 0
3 years ago
Plaskett's binary system consists of two stars that revolve in a circular orbit about a center of mass midway between them. This
vova2212 [387]

Answer:

1.554\times 10^{32}\ \text{kg}

Explanation:

M = Mass of each star

T = Time period = 15.5 days

v = Orbital velocity = 230 km/s

G = Gravitational constant = 6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2

Radius of orbit is given by

R=\dfrac{vT}{2\pi}

We have the relation

\dfrac{Mv^2}{R}=\dfrac{GM^2}{(2R)^2}\\\Rightarrow M=\dfrac{4Rv^2}{G}\\\Rightarrow M=\dfrac{4\dfrac{vT}{2\pi}v^2}{G}\\\Rightarrow M=\dfrac{2v^3T}{\pi G}\\\Rightarrow M=\dfrac{2\times 230000^3\times 15.5\times 24\times 60\times 60}{\pi\times 6.674\times 10^{-11}}\\\Rightarrow M=1.554\times 10^{32}\ \text{kg}

The mass of each star is 1.554\times 10^{32}\ \text{kg}

6 0
3 years ago
2. For each pair of variables, identify which is the independent and which is the dependent variable. a. How much gas is left in
Korvikt [17]

For each pair Independent variable and the dependent variable is -

a. How much gas is left in the gas tank vs. how far the car has traveled.

  • Independent variable =  how far the car has traveled
  • dependent variable = How much gas is left in the gas tank

b. How much money you've spent vs. how much money is in your wallet.

  • Independent variable =  How much money you've spent
  • dependent variable = how much money is in your wallet.

c. How far a toy car traveled vs. how much time went by​

  • Independent variable =  how much time went by​
  • dependent variable = How far a toy car traveled

An independent variable in any experiment or research is a variable that is manipulated or changed in the experiment, this change leads to a direct effect on the dependent variable.

A dependent variable is a variable that is directly affected by the independent variable and it is the variable that is measured or tested in an experiment.

Thus,

a. How much gas is left in the gas tank vs. how far the car has traveled.

  • Independent variable =  how far the car has traveled
  • dependent variable = How much gas is left in the gas tank

b. How much money you've spent vs. how much money is in your wallet.

  • Independent variable =  How much money you've spent
  • dependent variable = how much money is in your wallet.

c. How far a toy car traveled vs. how much time went by​

  • Independent variable =  how much time went by​
  • dependent variable = How far a toy car traveled

Learn more about dependent variables:

brainly.com/question/1670595:

3 0
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6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres
sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

2ah=v_f^2-v_0^2

Where:

\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}

If we assume the initial velocity to be 0 we get:

2ah=v_f^2

The acceleration is the acceleration due to gravity:

2gh=v_f^2

Now, we take the square root to both sides:

\sqrt{2gh}=v_f

Now, we substitute the values:

\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f

solving the operations:

280\frac{m}{s}=v

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

F_d=\frac{1}{2}C\rho_{air}Av^2

Where:

\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

F_d=mg

Now, we determine the mass of the raindrop using the following formula:

m=\rho_{water}V

Where:

\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}

The volume is the volume of a sphere, therefore:

m=\rho_{water}(\frac{4}{3}\pi r^3)

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)

Solving the operations:

m=1.39\times10^{-5}kg

Now, we substitute the values in the formula for the drag force:

F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})

Solving the operations:

F_d=1.36\times10^{-4}N

Now, we substitute in the formula:

1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2

Now, we solve for the velocity:

\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2

Now, we substitute the values. We will use the area of a circle:

\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2

Substituting the radius:

\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2

Solving the operations:

70.67\frac{m^2}{s^2}=v^2

Now, we take the square root to both sides:

\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\  \\ 8.4\frac{m}{s}=v \\  \end{gathered}

Therefore, the velocity is 8.4 m/s

7 0
1 year ago
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