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3 years ago

7

Please complete it if you know the answer. "The active region of a transistor is for.........

1 answer:

zubka84 [21]3 years ago

5 0

Answer:

the active region is bound by cutoff region and saturation or power dissipation region.

Explanation:

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Two particles each of mass m and charge q are suspended by strings of length / from a common point. Find the angle e that each s

ozzi

Answer:

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

Explanation:

Let the length of the string is L.

Let T be the tension in the string.

Resolve the components of T.

As the charge q is in equilibrium.

T Sinθ = Fe ..... (1)

T Cosθ = mg .......(2)

Divide equation (1) by equation (2), we get

tan θ = Fe / mg

$tan\theta =\frac{\frac{kq^{2}}{AB^{2}}}{mg}$

$tan\theta =\frac{\frac{kq^{2}}{4L^{2}Sin^{\theta }}}}{mg}$

$tan\theta =\frac{kq^{2}}{4L^{2}Sin^{2}\theta \times mg}$

$tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}$

As θ is very small, so tanθ and Sinθ is equal to θ.

$\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}$

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

7 0

2 years ago

If a body having mass 40kg started moving initially with rest and it takes a velocity of 20m/sec in time 4 seconds. Find the val

baherus [9]

${\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Mass \ of \ the \ body \ (m) = 40 \ kg}$

$\:\:\:\:\bullet\:\:\:\sf{Final \ velocity \ of \ the \ body \ (v) = 20 \ m/s}$

$\:\:\:\:\bullet\:\:\:\sf{Initial \ velocity \ of \ the \ body \ (u) = 0}$

$\\$

${\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\$

$\:\:\:\:\bullet\:\:\:\sf{Force \ exerted \ by \ the \ body \ ( F)}$

$\\$

${\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\$

☯ <u>Using 1st equation of motion </u>

$\\$

$\dashrightarrow\:\: \sf{v = u + at}$

$\\$

$\dashrightarrow\:\: \sf{20 = 0 + a(4)}$

$\\$

$\dashrightarrow\:\: \sf{20 = 4a}$

$\\$

$\dashrightarrow\:\: \sf{\dfrac{\cancel{20}}{\cancel{4}} = a}$

$\\$

$\dashrightarrow\:\: \sf{a = 5}$

$\\$

☯ <u>Now, Finding the force exerted </u>

$\\$

$\dashrightarrow\:\: \sf{F = ma}$

$\\$

$\dashrightarrow\:\: \sf{F = 40 \times 5}$

$\\$

$\dashrightarrow\:\: \sf{F = 200 \ N}$

$\\$

☯ <u>Hence</u>, $\\$

$\:\:\:\:\star\:\:\:\sf{The \ force \ exerted \ by \ the \ body \ is \ 200N}$

8 0

2 years ago

You want to estimate the diameter of a very small circular pinhole that you've made in a piece of aluminum foil. to do so, you s

Leya [2.2K]

Central maximum = d* wavelength/ D
thus
12*10-^3 = 3.4*6.32*10-^7/D
D = 3.4*6.32*10-^7/12*10-^3
D = 1.79*10-^4 m

3 0

3 years ago

An office window has dimensions 3.1 m by 2.1 m. As a result of the passage of a storm, the outside air pressure drops to 0.954 a

Virty [35]

Answer:

Net forces which pushes the window is 30342.78 N.

Explanation:

Given:

Dimension of the office window.

Length of the window = $3.1$ m

Width of the window = $2.1$ m

Area of the window = $(3.1\times 2.1) = 6.51\ m^2$

Difference in air pressure = Inside pressure - Outside pressure

= $(1.0-0.954)$ atm = $0.046$ atm

Conversion of the pressure in its SI unit.

⇒ $1$ atm = $101325$ Pa

⇒ $0.046$ atm = $0.046\times 101325 =4660.95$ Pa

We have to find the net force.

We know,

⇒ Pressure = Force/Area

⇒ $Pressure=\frac{Force }{Area}$

⇒ $Force =Pressure\times Area$

⇒ Plugging the values.

⇒ $Force =4660.95\times 6.51$

⇒ $Force=30342.78$ Newton (N)

So,

The net forces which pushes the window is 30342.78 N.

3 0

3 years ago

A 5 kg ball is suspended on one cable. Calculate the tension in the cable

ludmilkaskok [199]

Answer:

49N

Explanation:

F=ma

We know the mass is 5kg, and since the ball is suspended on one cable, the acceleration is g, 9.8m/s^2

F=5kg*9.8m/s^2

= 49N

Hope this helps!

7 0

2 years ago