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zubka84 [21]
2 years ago
8

Temperature of the resistor in an RC circuit. In this problem we will look at the temperature response of the resist

Physics
1 answer:
fomenos2 years ago
4 0

RC circuit determines the capacitor's charging rate.

  • In RC (resistive and capacitive) circuits, a capacitor's time constant is the number of seconds required to charge it to 63.2% of the input voltage.
  • This duration is described by a single time constant. After two time constants, the capacitor will be charged to 86.5% of the input voltage.
  • The RC time constant, also referred to as tau, is the time constant (in seconds) of an RC circuit and is obtained by multiplying the circuit resistance (in ohms) by the circuit capacitance (in farads), This transient reaction time T is stated in terms of = R x C, where R is the resistor value in ohms and C is the capacitor value in farads.

Learn more about  RC circuits here brainly.com/question/13450553

#SPJ4.

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I need help on question 6
Sauron [17]

I think that it’s letter C

4 0
4 years ago
You dip a wire loop into soapy water (n = 1.33) and hold it up vertically to look at the soap film in white light. The soap film
Gnesinka [82]

Answer:

the thickness of the soap film is 127.82 nm or 130 nm

Explanation:

Given the data in the question;

n = 1.33

λ_t = 680 nm = 680 × 10⁻⁹ m

m = 1 and β = 0

When we see the red fringe, its a point of maximum reflection

hence, for interference with a thin soap film, we say;

2 × n × d × cos( β ) = ( m - 0.5) × λ_t

so we substitute in our given values;

2 × 1.33 × d × cos( 0 ) = ( 1 - 0.5) × ( 680 × 10⁻⁹ )

2.66 × cos( 0 ) × d = 0.5 × ( 680 × 10⁻⁹ )

2.66 × 1 × d = 3.4 × 10⁻⁷

d = ( 3.4 × 10⁻⁷ ) / 2.66

d = 127.82 × 10⁻⁹ m

d = 127.82 nm ≈ 130 nm

Therefore, the thickness of the soap film is 127.82 nm or 130 nm

4 0
3 years ago
What are Earth's mineral resources and how long might<br> reserves last?
amm1812
Reserves might last 53.3 years.

Earth’s mineral resources:

3 0
3 years ago
in terms of the wavelength of the sound wave, about how far below the open end of the resonance tube is the first resonant posit
notka56 [123]

The first resonant position below the open end of the resonance tube is; <em><u>one-quarter of the wavelength</u></em>

In the event of the first resonant position in a resonance tube, there will be a maximum air displacement which is only one antinode right at the open end where the motion is constrained.

However, there will be no displacement at the closed end which means another one node right at the closed end where air is halted.

This means that the standing wave will have one-quarter of the wavelength in the test tube.

Thus;

L = ¼λ

Read more at; brainly.com/question/17086525

8 0
2 years ago
A small object with a 5.0-mC charge is accelerating horizontally on a friction-free surface at 0.0050 m/s2 due only to an electr
kolbaska11 [484]

Answer:

0.002 N/C

Explanation:

Parameters given:

Charge of object, q = 5 mC = 5 * 10^{-3} C

Acceleration of object, a = 0.005 m/s^2

Mass of object, m = 2.0 g

The Electric field exerts a particular force on the object, causing it to accelerate (Electrostatic force).

We know that Electrostatic force, F, is given in terms of Electric field, E, as:

F = qE

This means that the object exerts a force of -qE on the Electric force (Action with equal and opposite reaction).

The object also has a force, F, due to its acceleration a. This force is the product of its mass and acceleration. Mathematically:

F = ma

Equating the two forces of the object, we get:

-qE = ma

=> E = \frac{-ma}{q}

Solving for E, we have:

E = \frac{-2 * 10^{-3} * 0.005}{5 * 10^{-3}} \\\\\\E = -0.002 N/C

The magnitude will be:

|E| = |-0.002| N/C = 0.002 N/C

The electric field has a magnitude of 0.002 N/C.

4 0
3 years ago
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