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3 years ago

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

Physics

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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Water flows over a section of Niagara Falls at rate of 1.1×10^6 kg/s and falls 49.4 m. How much power is generated by the fallin

DerKrebs [107]

Answer:

The power generated is = 5.33×10⁸ Watt.

Explanation:

Power: Power can be defined as the time rate of doing work. The S.I unit of power is Watt(W).

Mathematically,

Power (P) = Work done/time or Energy/time

P = mgh/t............................... Equation 1

P = δgh............................. Equation 2

Where δ = fall rate, g = acceleration due to gravity, h = height.

Given: δ = 1.1×10⁶ kg/s, h = 49.4 m g = 9.81 m/s²

Substituting these values into equation 2

P = 1.1×10⁶×49.4×9.81

P = 533.08×10⁶

P = 5.33×10⁸ Watt.

Thus the power generated is = 5.33×10⁸ Watt.

5 0

3 years ago

An electron has an initial velocity of (19.0 j + 18.0 k) km/s and a constant acceleration of (3.00 ✕ 1012 m/s2)i in a region in

asambeis [7]

Answer:

$E=(-17.08 i +7.2 j -7.6 k )N/C$

Explanation:

$v= (19.0j+18.k)km/s$

$a=3.0x10^{13}m/s^2$

$\beta= 400x10{-6}T$

Electron information needed to solve the question:

$m_e=9.11x10^{-31}kg$

$q=-1.6x10{-19}C$

$F=F_E+F_B=q*(E+Vx \beta)$

$F=m*a$

$m*a=q*(E+Vx\beta)$

$E=\frac{m*a}{q}-(Vx\beta )$

$E=\frac{9.11x10{-31}kg*3.0x10^{12}m/s^2}{-1.6x10{-19}C}-[(19.0x10^3mj+18.0x10^3m)xi(400x10^{-6}T)]$

$E=-i17.08N/C-[7.6(-k)+7.2(j)]N/C$

$E=(-17.08 i +7.2 j -7.6 k )N/C$

6 0

3 years ago

A uniform-density 7 kg disk of radius 0.21 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is

GREYUIT [131]

Answer:

$\omega = 22.67 rad/s$

Explanation:

Here we can use energy conservation

As per energy conservation conditions we know that work done by external source is converted into kinetic energy of the disc

Now we have

$W = \frac{1}{2}I\omega^2$

now we know that work done is product of force and displacement

so here we have

$W = F.d$

$W = (44 N)(0.9 m) = 39.6 J$

now for moment of inertia of the disc we will have

$I = \frac{1}{2}mR^2$

$I = \frac{1}{2}(7 kg)(0.21^2)$

$I = 0.154 kg m^2$

now from above equation we will have

$39.6 = \frac{1}{2}(0.154)\omega^2$

$\omega = 22.67 rad/s$

5 0

3 years ago

Read 2 more answers

Two electrons with a charge of magnitude 1.6×10-19 C in an atom are separated by 1.5×10-10 m, the typical size of an atom. What

vesna_86 [32]

Answer:

$1.02\cdot 10^{-8} N$ , repulsive

Explanation:

The magnitude of the electric force between two charged particles is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges of the two particles

r is the separation between the two charges

The force is:

- repulsive if the two charges have same sign

- Attractive if the two charges have opposite signs

In this problem, we have two electrons, so:

$q_1=q_2=1.6\cdot 10^{-19}C$ is the magnitude of the two electrons

$r=1.5\cdot 10^{-10} m$ is their separation

Substituting into the formula, we find the electric force between them:

$F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(1.5\cdot 10^{-10})^2}=1.02\cdot 10^{-8} N$

And the force is repulsive, since the two electrons have same sign charge.

4 0

3 years ago

Describe what happens to the electric field lines when two objects with unlike charges are brought near each other.

harina [27]

Hello.

The answer is:

It creates a spark.

Then thespark can sometimes start a fire or other serious problems.

Have a nice day

4 0

3 years ago

Read 2 more answers