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ololo11 [35]
3 years ago
13

A beam of EMW in air strikes a sheet of a glass at 75 degrees with the normal in air. You observe that red light makes an angle

of 38.1 degrees with the normal in the glass, while violet light makes a 36.7 degree angle.
1. What are the indexes of refraction of this glass for these colors of light?
Answer in the order indicated. Separate your answers with a comma.
2. What are the speeds of red and violet light in the glass?
Physics
1 answer:
RUDIKE [14]3 years ago
5 0

Answer:

a) 1.57, 1.61

B 1.91x10^8m/s , 1.86 x 10^8m/s

Explanation:

Ok we know that refractive index is given as = sin i/ sin r

And i is the angle of incidence in the air and r is angle of refraction in the glass,

So we can sa y

for red light n = sin75/sin38.1 = 1.57

and for violet, n =sin75/sin36.7 = 1.61

So

(a) 1.57, 1.61

(b) we know that

Refractive index n = c/v,

where c is speed off light in air and v is speed of light in glass,

So , v = c/n

for red light v = c/1.57 = 1.91 x 10^8 m/s

and for violet light, v = c/1.61 =

1.86 x 10^8 m/s

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Answer:

Answer:

28.025 Nm

Explanation:

Angular acceleration, α = 29.5 rad/s^2

oment of inertia, I = 0.95 kg m^2

The torque is defined as

τ = I x α

τ = 0.95 x 29.5

τ = 28.025 Nm

Thus, the torque is 28.025 Nm.

Explanation:

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Which of these galaxies would you most likely find at the center of a large cluster of galaxies?
Arada [10]

Answer:

b. a large elliptical galaxy

Explanation:

In elliptical galaxies the stars are grouped in an elliptical shape, it has a low quantity of gas and dust in comparison to spiral galaxies, and its stars belong to an old population, there is not new stellar formation in it.

The stars orbit in a messy way which made to believe that they form from the merger of galaxies.

They are also really massive (around 10^{4} solar masses).

The most massive and luminous can be found in the center of cluster of galaxies.  

4 0
3 years ago
What is the mass, in kilograms, of an avogadro's number of people, if the average mass of a person is 170 lb ?
andre [41]
About 4.7e25 kilograms
8 0
3 years ago
A 3.0-A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the
harina [27]

Answer:

R=2.78\ \Omega

Explanation:

Given that,

The current flowing in the circuit, I = 3 A

The power of the battery, P = 25 W

We need to find the resistance of the battery. We know that the power of the battery is given by the formula as follows :

P=I^2R

Put all the values to find R.

R=\dfrac{P}{I^2}\\\\R=\dfrac{25}{(3)^2}\\\\R=2.78\ \Omega

So, the resistance is equal to 2.78\ \Omega.

7 0
2 years ago
An alternative to CFL bulbs and incandescent bulbs are light-emitting diode (LED) bulbs. A 16-W LED bulb can replace a 100-W inc
Yanka [14]

Answer:

LED bulb = 0.145 A

Incandescent bulb = 0.909 A

CFL bulb = 0.218 A

Explanation:

Given:

Power rating of LED bulb (P₁) = 16 W

Power rating of incandescent bulb (P₂) = 100 W

Power rating of CFL bulb (P₃) = 24 W

Terminal voltage across the circuit (V) = 110 V

We know that, power is related to terminal voltage and current drawn as:

P=VI

Express this in terms of 'I'. This gives,

I=\frac{P}{V}

Now, calculate the current drawn in each bulb using their respective values.

For LED bulb, P_1=16\ W, V=110\ V

So, current drawn is given as:

I_1=\frac{16\ W}{110\ V}=0.145\ A

For incandescent bulb, P_2=100\ W, V=110\ V

So, current drawn is given as:

I_2=\frac{100\ W}{110\ V}=0.909\ A

For CFL bulb, P_3=24\ W, V=110\ V

So, current drawn is given as:

I_3=\frac{24\ W}{110\ V}=0.218\ A

Therefore, the currents drawn through LED bulb, incandescent bulb and CFL bulb are 0.145 A, 0.909 A and 0.218 A respectively.

5 0
3 years ago
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