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yan [13]
2 years ago
13

A single light beam is split into two equal beams, denoted a and b. beam a travels through a medium with a higher index of refra

ction than the medium that beam b travel through. when both beams exit their media back into air, how do their wavelengths compare
Physics
1 answer:
nevsk [136]2 years ago
4 0

Answer:

c) Both beams have the same wavelength.

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a desktop computer and monitor together draw about 2 A of current they plug into a wall outlet that is 120 V what is the Resista
Delvig [45]

Answer:

60 \Omega

Explanation:

the relation between current, voltage and resistance in an electrical circuit is given by Ohm's law:

V=IR

where V is the voltage, I is the current and R is the resistance. In this problem, the current is I=2 A, the voltage is V=120 V, therefore we can arrange the previous equation and find the resistance:

R=\frac{V}{I}=\frac{120 V}{2 A}=60 \Omega

7 0
3 years ago
Which of the following is most useful to determine how much energy is being used by a circuit in a given amount of time?
user100 [1]

Answer:

The answer is A.

Explanation:

5 0
3 years ago
A 60-W, 120-V light bulb and a 200-W, 120-V light bulb are connected in series across a 240-V line. Assume that the resistance o
gulaghasi [49]

A. 0.77 A

Using the relationship:

P=\frac{V^2}{R}

where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.

For the first light bulb, P = 60 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{60 W}=240 \Omega

For the second light bulb, P = 200 W and V = 120 V, so the resistance is

R_1=\frac{V^2}{P}=\frac{(120 V)^2}{200 W}=72 \Omega

The two light bulbs are connected in series, so their equivalent resistance is

R=R_1 + R_2 = 240 \Omega + 72 \Omega =312 \Omega

The two light bulbs are connected to a voltage of

V  = 240 V

So we can find the current through the two bulbs by using Ohm's law:

I=\frac{V}{R}=\frac{240 V}{312 \Omega}=0.77 A

B. 142.3 W

The power dissipated in the first bulb is given by:

P_1=I^2 R_1

where

I = 0.77 A is the current

R_1 = 240 \Omega is the resistance of the bulb

Substituting numbers, we get

P_1 = (0.77 A)^2 (240 \Omega)=142.3 W

C. 42.7 W

The power dissipated in the second bulb is given by:

P_2=I^2 R_2

where

I = 0.77 A is the current

R_2 = 72 \Omega is the resistance of the bulb

Substituting numbers, we get

P_2 = (0.77 A)^2 (72 \Omega)=42.7 W

D. The 60-W bulb burns out very quickly

The power dissipated by the resistance of each light bulb is equal to:

P=\frac{E}{t}

where

E is the amount of energy dissipated

t is the time interval

From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.

7 0
3 years ago
which of the following cannot be increased by using a machine of some kind? work, force, speed, torque
Lemur [1.5K]

Explanation:

Work cannot be increased by using a machine of some kind.

8 0
3 years ago
Light behaves like a) a wave at times, and a particle at other times.
Damm [24]

The wave-particle dual nature of light has been documented and tested many times.

Choice A

6 0
3 years ago
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