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3 years ago

Help with these questions please?

Physics

1 answer:

egoroff_w [7]3 years ago

8 0

Answer:

5. dispersion

6. 49.8°

Explanation:

5. Dispersion is the name given to the phenomenon of light of different wavelengths being bent differently. A rainbow is the result of light from a point source (the sun) being spread out by wavelength (color), a nice example of dispersion.

___

6. n = 1.31 is the ratio of the sine of the angle of refraction to the sine of the angle of incidence (for light passing to a medium of n = 1). When the angle of refraction is 90°, the angle of incidence is the "critical angle." So, ...

sin(90°)/sin(critical) = 1.31

critical angle = arcsin(1/1.31) ≈ 49.8°

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A series circuit has a voltage supply of 12 V and a resistor of 2.4 kΩ. How much power is dissipated by the resistor?

adell [148]

Answer:

(B) 0.06W

Explanation:

power absorbed by the resistor is given by $\frac{V^2}{R}=I^2R$

Where V = voltage

R= resistance

I = current through the circuit

we have given V =12 Volt and resistance =2.4 K $\Omega$

current $I=\frac{V}{R}=\frac{12}{2.4\times 1000}=5\times 10^{-3}A=5mA$

power using voltage and resistance equation

$p=\frac{12^2}{2.4\times 1000}=60mW$ =0.06W

using current equation $P=I^2R=5^2\times 12=60mW$ = 0.06W

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3 years ago

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Ludmilka [50]

Answer:

For the first one, its "attract"

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Give you a device that tracks your speed?

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3 years ago

Which of the following best describes the use of a renewable resource? (2 points) Most power plants burn fossil fuels to generat

kolbaska11 [484]

Which of the following best describes the use of a renewable resource?

Answer:

There are areas in California that use the heat from deep in the Earth to generate electricity.

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3 years ago

A 15.0 Ohms resistor is connected in series to a 120V generator and two 10.0 Ohms resistors that are connected in parallel to ea

Nostrana [21]

Hi there! :)

Reference the diagram below for clarification.

We must begin by knowing the following rules for resistors in series and parallel.

In series:
$R_T = R_1 + R_2 + ... + R_n$

In parallel:
$\frac{1}{R_T} = \frac{1}{R_1} + \frac{1}{R_2} + ... + \frac{1}{R_n}$

We can begin solving for the equivalent resistance of the two resistors in parallel using the parallel rules.

$\frac{1}{R_{T, parallel}} = \frac{1}{10} + \frac{1}{10}\\\\\frac{1}{R_{T, parallel}} = \frac{2}{10} = \frac{1}{5}\\\\R_{T, parallel} = 5\Omega$

Now that we have reduced the parallel resistors to a 'single' resistor, we can add their equivalent resistance with the other resistor in parallel (15 Ohm) using series rules:
$R_T = 15 + 5\\\\\boxed{R_T = 20 \Omega}$

We can use Ohm's law to solve for the current in the circuit.

$i = \frac{V}{R_T}\\\\i = \frac{120}{20} = \boxed{6 A}$

For resistors in series, both resistors receive the SAME current.

Therefore, the 15Ω resistor receives 6A, and the parallel COMBO (not each individual resistor, but the 5Ω equivalent when combined) receives 6A.

In this instance, since both of the resistors in parallel are equal, the current is SPLIT EQUALLY between the two. (Current in parallel ADDS UP). Therefore, an even split between 2 resistors of 6 A is <u>3A for each 10Ω resistor</u>.

Since the 15.0 Ω resistor receives 6A, we can use Ohm's Law to solve for voltage.

$V = iR\\\\V = (6)(15) = \boxed{90 V}$

4 0

2 years ago