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vampirchik [111]
3 years ago
7

While teaching about transistors, Mr. Mendoza shows a video of a hand pump being used to draw water from a well. He points out h

ow the person primes the pump by pouring a small amount of water into it before the water would start to flow. No matter how much water is used to prime the pump, once started, the pump provides a steady flow of water while the handle is operated.
What is Mr. Mendoza using the hand pump to model?

A. its use as a switch to make a small signal into a large signal
B. its use as an amplifier to make a small signal into a large signal
C. its use as a switch to turn on a large current using a small current
D. its use as an amplifier to turn on a large current using a small current
Physics
2 answers:
Alchen [17]3 years ago
5 0

Answer:

The answer is B,  Mr Mendoza is using the hand pump to model an amplifier to make a small signal into a large signal.

Explanation:

An amplifier basically is an electronic gadget used to cover lower signal or current  input into higher current, signal or output. In a situation where water is being pumped into a reservoir and the level of the reservoir is higher than that of the well, am amplifier can used used with the pump to increase the output current going into the pump thereby increase the speed and flow of water into the reservoir.

storchak [24]3 years ago
5 0

Answer:

B

Explanation:

Transistors are small pieces of semiconductor electronic devices that are used basically for amplification and switching of signals.

They are used to amplify a small signal to a larger one.

They function as switches as they determine if current flows through the collector emitter path. When a voltage is applied to the base it allows currents to flow through to C-E path, but if no current is applied no current will flow .

In the illustration given in the question the pump provides a steady flow of water when a small quantity of water is fed into it. This can be compared to the use of transistors as an amplifier in converting small signals into large signals.

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alexandr1967 [171]
An equilibruium is not changed by a changed in pressure. the answer is false
7 0
3 years ago
The magnitude of the force associated with the gravitational field is constant and has a value FF . A particle is launched from
uysha [10]

Answer:

The kinetic energy of the particle will be 12U₀

Explanation:

Given that,

A particle is launched from point B with an initial velocity and reaches point A having gained U₀ joules of kinetic energy.

Constant force = 12F

According to question,

The kinetic energy is

U_{0}=Fx....(I)

Constant force = 12F

A resistive force field is now set up ,

Resistive force is given by,

F_{r}=12F

When the particle moves from point B to point A then,

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Using formula for kinetic energy

U=F_{r}x

Put the value of F_{r}

U=12Fx

Now, from equation (I)

U=12U_{0}

Hence, The kinetic energy of the particle will be 12U₀.

7 0
3 years ago
A projectile is launched at an angle above the
gtnhenbr [62]
The first rule of vectors is that the horizontal and vertical components are separate. Disregarding air resistance, the only thing we have to worry about is gravity.

The appropriate suvat to use for the vertical component is v = u +at
I will take a to be -9.81, you may have to change it to be 10 if your qualification likes g to be 10.

v = 30 + (-9.81x2)
v = 30 - 19.62
=10.38m/s

Therefore we know that after 2.0 s the vertical component will be 10.38ms^-1, ie 10m/s as the answers given are all to 2sf.

The horizontal component is completely separate to the vertical component and since there is no air resistance, it will remain constant throughout the projectiles trajectory. Therefore it will remain at 40ms^-1.

Combining this together we get:
(1) vx=40m/s and vy=10m/s

7 0
3 years ago
A mass is attached to a vertical spring, which then goes into oscillation. At the high point of the oscillation, the spring is i
andrew-mc [135]

Answer:

0.34 sec

Explanation:

Low point of spring ( length of stretched spring ) = 5.8 cm

midpoint of spring = 5.8 / 2 = 2.9 cm

Determine the oscillation period

at equilibrum condition

Kx = Mg

g= 9.8 m/s^2

x = 2.9 * 10^-2 m

k / m = 9.8 / ( 2.9 * 10^-2 ) =  337.93

note : w = \sqrt{k/m}   = \sqrt{337.93} = 18.38 rad/sec

Period of oscillation =  2\pi  / w

                                  = 0.34 sec

8 0
3 years ago
What is the technology used behind scanning probe microscopes?
Wittaler [7]

hola ya sabes  la respuesta si si ,me la dices porfa

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3 years ago
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