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Alex17521 [72]
3 years ago
5

Which solid-state component can be used as a switch to turn current on or off?

Physics
2 answers:
geniusboy [140]3 years ago
7 0

Answer:

Transistor

Explanation:

Papessa [141]3 years ago
4 0
The answer is Transistor. Its a semiconductor device used to amplify or switch electronic signals and electrical power. It is composed of semiconductor material usually with at least three terminals for connection to an external circuit.
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telo118 [61]

A = 4\pi r^2

A = 4\pi (2\mu m /2)^2 (10^{-6}m/1\mu m)^2 (1mm/10{-3})^2

A = 1.33*!0^{-5}MM^2

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The formula K2S indicates that
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There are two atoms of potassium bonded to one atom of sulfur.
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A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 50.9 c
seraphim [82]

Answer:

Flow rate 2.34 m3/s

Diameter 0.754 m

Explanation:

Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.

The area at the well head is

A = \pi r_w^2 = \pi (0.509/2)^2 = 0.203 m^2

So the volume flow rate along the pipe is

\dot{V} = Av = 0.203 * 11.5 = 2.34 m^3/s

We can use the similar logic to find the cross-section area at the refinery

A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2

The radius of the pipe at the refinery is:

A_r = \pi r^2

r^2 =A_r/\pi = 0.446/\pi = 0.141

r = \sqrt{0.141} = 0.377m

So the diameter is twice the radius = 0.38*2 = 0.754m

6 0
3 years ago
Scientists have proven that genes play no role in self-esteem. Please select the best answer from the choices provided. T F
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3 years ago
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.1 An 8-ft 3 tank contains air at an initial temperature of 808F and initial pressure of 100 lbf/in. 2 The tank develops a small
Alina [70]

Correct temperature is 80°F

Answer:

T_f = 38.83°F

Explanation:

We are given;

Volume; V = 8 ft³

Initial Pressure; P_i = 100 lbf/in² = 100 × 12² lbf/ft²

Initial temperature; T_i = 80°F = 539.67 °R

Time for outlet flow; t_o = 90 s

Mass flow rate at outlet; m'_o = 0.03 lb/s

Final pressure; P_f = 30 lbf/in² = 30 × 12² lbf/ft²

Now, from ideal gas equation,

Pv = RT

Where v is initial specific volume

R is ideal gas constant = 53.33 ft.lbf/°R

Thus;

v = RT/P

v_i = 53.33 × 539.67/(100 × 12²)

v_i = 2 ft³/lb

Formula for initial mass is;

m_i = V/v_i

m_i = 8/2

m_i = 4 lb

Now change in mass is given as;

Δm = m'_o × t_o

Δm = 0.03 × 90

Δm = 2.7 lb

Now,

m_f = m_i - Δm

Thus; m_f = 4 - 2.7

m_f = 1.3 lb

Similarly in above;

v_f = V/m_f

v_f = 8/1.3

v_f = 6.154 ft³/lb

Again;

Pv = RT

Thus;

T_f = P_f•v_f/R

T_f = (30 × 12² × 6.154)/53.33

T_f = 498.5°R

Converting to °F gives;

T_f = 38.83°F

7 0
3 years ago
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