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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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What net force would be needed to accelerate the box at 2.3 m/s^2
Akimi4 [234]
The pertinent equation here is F=ma.  You haven't shared the mass of the box, so I will use M to represent that mass.

Then F = M(<span>2.3 m/s^2)        (answer)</span>
6 0
3 years ago
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a cyclist coasting down a 5.0 ◦ incline at a constant speed of 6.0 km/h because of air resistance. If the total mass of the bicy
Dvinal [7]

Answer:

F_{net}= 85.41\ N

Explanation:

mass of the bicycle + cyclist = 50 kg

constant speed = 6 km/h

a cyclist coasting down a 5.0° incline

the downward velocity is constant, so net acceleration must be zero

the air drag must be equal to gravitational force downward along the ramp

F_a = mg sin \theta  

now for upward motion

F_{net} = mg sin \theta + air\ drag

F_{net} = mg sin \theta + mg sin \theta

F_{net} = 2 mg sin \theta

F_{net} = 2\times 50 \times 9.8 sin 5^0

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3 0
3 years ago
What is the frequency of a wave with a wavelength of 15 m and a wavespeed of<br> 300 m/s?
STatiana [176]

Answer: f=20 (i think)

Explanation:

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6 0
3 years ago
Complete the sentence-
trasher [3.6K]

Answer:

Friction always acts opposite to the motion.

3 0
2 years ago
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A uranium nucleus is traveling at 0.94 c in the positive direction relative to the laboratory when it suddenly splits into two p
Anettt [7]

Answer:

A   u = 0.36c      B u = 0.961c

Explanation:

In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains

     u ’= (u-v) / (1- uv / c²)

Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory

The data give is u ’= 0.43c and the initial core velocity v = 0.94c

Let's clear the speed with respect to the observer (u)

      u’ (1- u v / c²) = u -v

      u + u ’uv / c² = v - u’

      u (1 + u ’v / c²) = v - u’

      u = (v-u ’) / (1+ u’ v / c²)

Let's calculate

      u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)

      u = 0.51c / (1 + 0.4042)

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We repeat the calculation for the other piece

In this case u ’= - 0.35c

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       u = 1.29c / (1- 0.329)

       u = 0.961c

6 0
3 years ago
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