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2 years ago

What’s a pnp transitor?

Engineering

2 answers:

Dafna1 [17]2 years ago

7 0

Answer:

A transistor is of two types: PNP and NPN

Explanation:

A transistor is current controlled device in which the current flow by the movement of the electrons and wholes.

The NPN and PNP transistor are the types of the BJT (Bipolar junction transistor) that have two uses i.e switching and amplification.

A PNP transistor have three junction i.e P, N and P where the base is connected to the N - junction. The P junction is high in number of wholes and N- junction have high numbers of electrons.

In the PNP transistor the voltage is provided to the emitter and base terminal with base connected to the negative terminal of the voltage source and emitter connected to its positive terminal. The direction of the flow of the current is from emitter to collector whenever a significant voltage is provided to the emitter base junction.

Damm [24]2 years ago

6 0

Answer:

PnP transistor is a current controlled device. It has two crystal diodes connected back to back. The left side of the diode is known as the emitter-base diode and the other side which is the right side of the diode is known as the collector-base diode.

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An engine has been diagnosed with blowby.

12345 [234]

B, tech b only, Blow by is describe or defined as air going by the compression rings. Hope it helps

8 0

2 years ago

What is the relative % change in P if we double the absolute temperature of an ideal gas keeping mass and volume constant?

Contact [7]

Answer: 100% (double)

Explanation:

The question tells us two important things:

Mass remains constant
Volume remains constant

(We can think in a gas enclosed in a closed bottle, which is heated, for instance)

In this case we know that, as always the gas can be considered as ideal, we can apply the general equation for ideal gases, as follows:

State 1 (P1, V1, n1, T1) ⇒ P1*V1 = n1*R*T1
State 2 (P2, V2, n2, T2) ⇒ P2*V2 = n2*R*T2

But we know that V1=V2 and that n1=n2, som dividing both sides, we get:

P1/P2 = T1/T2, i.e, if T2=2 T1, in order to keep both sides equal, we need that P2= 2 P1.

This result is just reasonable, because as temperature measures the kinetic energy of the gas molecules, if temperature increases, the kinetic energy will also increase, and consequently, the frequency of collisions of the molecules (which is the pressure) will also increase in the same proportion.

6 0

3 years ago

An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,

tester [92]

Answer:

percentage change in volume is 2.60%

water level rise is 4.138 mm

Explanation:

given data

volume of water V = 500 L

temperature T1 = 20°C

temperature T2 = 80°C

vat diameter = 2 m

to find out

percentage change in volume and how much water level rise

solution

we will apply here bulk modulus equation that is ratio of change in pressure to rate of change of volume to change of pressure

and we know that is also in term of change in density also

E = $-\frac{dp}{dV/V}$ ................1

And $-\frac{dV}{V} = \frac{d\rho}{\rho}$ ............2

here ρ is density

and we know ρ for 20°C = 998 kg/m³

and ρ for 80°C = 972 kg/m³

so from equation 2 put all value

$-\frac{dV}{V} = \frac{d\rho}{\rho}$

$-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}$

dV = 0.0130 m³

so now % change in volume will be

dV % = $-\frac{dV}{V}$ × 100

dV % = $-\frac{0.0130}{500*10^{-3} }$ × 100

dV % = 2.60 %

so percentage change in volume is 2.60%

and

initial volume v1 = $\frac{\pi }{4} *d^2*l(i)$ ................3

final volume v2 = $\frac{\pi }{4} *d^2*l(f)$ ................4

now from equation 3 and 4 , subtract v1 by v2

v2 - v1 = $\frac{\pi }{4} *d^2*(l(f)-l(i))$

dV = $\frac{\pi }{4} *d^2*dl$

put here all value

0.0130 = $\frac{\pi }{4} *2^2*dl$

dl = 0.004138 m

so water level rise is 4.138 mm

8 0

2 years ago

3 examples of technology transfer pls

OLEGan [10]

Answer: electrical, mathematical, and geographical

Explanation: Yee

- Cash Nasty

3 0

2 years ago

Read 2 more answers

A spring-loaded piston-cylinder contains 1 kg of carbon dioxide. This system is heated from 104 kPa and 25 °C to 1,068 kPa and 3

labwork [276]

Answer:

Q = -68.859 kJ

Explanation:

given details

mass $co_2 = 1 kg$

initial pressure P_1 = 104 kPa

Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K

final pressure P_2 = 1068 kPa

Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K

we know that

molecular mass of $co_2 = 44$

R = 8.314/44 = 0.189 kJ/kg K

c_v = 0.657 kJ/kgK

from ideal gas equation

PV =mRT

$V_1 = \frac{m RT_1}{P_1}$

$=\frac{1*0.189*298}{104}$

$V_1 = 0.5415 m3$

$V_2 = \frac{m RT_2}{P_2}$

$=\frac{1*0.189*584}{1068}$

$V_1 = 0.1033 m3$

WORK DONE

$W =P_{avg}*{V_2-V_1}$

w = 586*(0.1033 -0.514)

W =256.76 kJ

INTERNAL ENERGY IS

$\Delta U = m *c_v*{V_2-V_1}$

$\Delta U = 1*0.657*(584-298)$

$\Delta U =187.902 kJ$

HEAT TRANSFER

$Q = \Delta U +W$

= 187.902 +(-256.46)

Q = -68.859 kJ

7 0

3 years ago