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san4es73 [151]
3 years ago
14

What is a good analogy for potential energy in a circuit?

Physics
1 answer:
AlexFokin [52]3 years ago
8 0

Answer: find the answer in the explanation

Explanation: The potential energy in a circuit between two points is the workdone in taking units positive charge from one point to the other in the circuit. Its S.I unit is Joule.

Because voltage of a battery or cell in a circuit is the potential energy difference across its terminal

The potential energy difference between two points is one volt, if the workdone in taking one Coulomb of positive charge from one point to the other is one Joule.

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Acid rain is the direct result of acid released by burning fossil fuels. T/F
disa [49]

Answer : False.

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3 years ago
A very long string (linear density 0.7 kg/m ) is stretched with a tension of 70 N . One end of the string oscillates up and down
rewona [7]

To develop this problem it is necessary to apply the concepts related to Wavelength, The relationship between speed, voltage and linear density as well as frequency. By definition the speed as a function of the tension and the linear density is given by

V = \sqrt{\frac{T}{\rho}}

Where,

T = Tension

\rho = Linear density

Our data are given by

Tension , T = 70 N

Linear density , \rho = 0.7 kg/m

Amplitude , A = 7 cm = 0.07 m

Period , t = 0.35 s

Replacing our values,

V = \sqrt{\frac{T}{\rho}}

V = \sqrt{\frac{70}{0.7}

V = 10m/s

Speed can also be expressed as

V = \lambda f

Re-arrange to find \lambda

\lambda = \frac{V}{f}

Where,

f = Frequency,

Which is also described in function of the Period as,

f = \frac{1}{T}

f = \frac{1}{0.35}

f = 2.86 Hz

Therefore replacing to find \lambda

\lambda = \frac{10}{2.86}

\lambda = 3.49m

Therefore the wavelength of the waves created in the string is 3.49m

3 0
3 years ago
The moon's surface gravity is one-sixth that of the earth. Calculate the weight on the moon of an object that has a mass of 24 k
ad-work [718]
When we say "<span>The moon's surface gravity is one-sixth that of the earth.",
we mean that the acceleration of gravity on the Moon's surface is 1/6 of
the acceleration of gravity on the Earth's surface.

The acceleration of gravity is (9.8 m/s</span>²) on the Earth's surface, so
<span>it would be (9.8/6 m/s</span>²) on the Moon's surface.
<span>
The weight of any object, right now, is

(object's mass) </span>· (acceleration of gravity where the object is located now) .
<span>
If the object's mass is 24 kg and the object is on the Moon right now,
then its weight is 

(24 kg) </span>· (9.8/6 m/s²)

= (24 · 9.8 / 6) kg-m/s²

= 39.2 Newtons
7 0
3 years ago
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