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3 years ago

Explain the necessity of smoothing the output voltage before applying it to a transistor

1 answer:

6 0

In simple words, smoothing can be defined as the circuit which is done to eliminate the ripple from the yield of a direct current power supply.

Explanation:

Necessity of smoothing:

Smoothing can be also applied in the form of a capacitor that acts to decrease or level out variations in a signal. And these capacitors are mostly used after power supply in voltage.
The yield DC voltage of a half-wave rectifier provided in the figure of a sinusoidal wave.
In a method to provide a constant DC voltage from a corrected AC source, a filter or smoothing circuit is required.

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A car moves at a constant speed of 10 m/s. If the car doesn't accelerate during the next 40 s how far will it go?

tangare [24]

the answer is b. space = time * velocity

8 0

3 years ago

2. What biotic and abiotic factors might influence the wolf and moose population numbers? List 1 biotic factors 1 abiotic factor

Ulleksa [173]

Answer:

Abiotic - sun Biotic- Plants

Water supply, climate, shape of the land, vegetation, soils and availability of natural resources.

Explanation:

Abiotic means non living so the sun is non living. The sun gives us warmth and the ability to survive

Biotic means alive or living- plants are living and give off oxygen and take in carbon to help us live umm ok

6 0

3 years ago

plates of a parallel-plate capacitor are 2.50 mm apart, and each carries a charge of magnitude 85.0 nC . The plates are in vacuu

tensa zangetsu [6.8K]

Answer:

12500 V

Explanation:

The electric field in the gap of a parallel-plate capacitor is uniform, so the following relationship between electric field strength, potential difference and distance can be used:

$\Delta V = E d$

where

$\Delta V$ is the potential difference between the plates

E is the electric field strength

d is the distance between the plates

For the capacitor in this problem, we have

$E=5.00\cdot 10^6 V/m$

$d = 2.50 mm = 2.50\cdot 10^{-3} m$

Substituting, we find

$\Delta V = (5.00\cdot 10^6)(2.50\cdot 10^{-3})=12500 V$

3 0

3 years ago

Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1

hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

$\sum F=ma$

We know that the force between two bodies due to gravity is given by the following equation:

$F_{g} = G\frac{m_{1}m_{2}}{r^{2}}$

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

$F_{g} =G\frac{Mm}{r^{2}}$

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

$F_{c}=ma_{c}$

where the centripetal acceleration is given by:

$a_{c}=\omega ^{2}r$

where

$\omega = \frac{2\pi}{T}$

Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

or:

$a_{c}=\frac{4\pi^{2}r}{T^{2}}$

so:

$F_{c}=m(\frac{4\pi^{2}r}{T^{2}})$

so now we can do the sum of forces:

$\sum F=ma$

$F_{g}=ma_{c}$

$G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})$

in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

we can now solve this for $T^{2}$ so we get:

$T^{2} = \frac{4\pi ^{2}r^{3}}{GM}$

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

$T_{1}^{2}>T_{2}^{2}$

So let's see what's going on there, we'll call:

$M_{1}$ = mass of Star 1

$M_{2}$ = mass of Star 2

So:

$\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}$

we can get rid of all the constants so we end up with:

$\frac{1}{M_{1}}>\frac{1}{M_{2}}$

and let's flip the inequality, so we get:

$M_{2}>M_{1}$

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0

3 years ago

What is the average acceleration given by this graph:

Zepler [3.9K]

Answer:

Explanation:

There is no graph.

5 0

3 years ago