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enyata [817]
3 years ago
7

What is the device used to measure a potential difference in a circuit called?

Physics
2 answers:
kvasek [131]3 years ago
8 0
A voltmeter is the instrument used to measure a potential difference between two points in an electric circuit
Romashka-Z-Leto [24]3 years ago
4 0
The potential difference between two points in a circuit is
sometimes called the 'voltage' because its unit of measure
is the volt.  This is a big part of the reason why a voltmeter
is the device that's used to measure it.
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Two ships leave a harbor at the same time, traveling on courses that have an angle of 110∘ between them. If the first ship trave
Allushta [10]

Answer:

49.07 miles

Explanation:

Angle between two ships = 110° = θ

First ship speed = 22 mph

Second ship speed = 34 mph

Distance covered by first ship after 1.2 hours = 22×1.2 = 26.4 miles = b

Distance covered by second ship after 1.2 hours = 34×1.2 = 40.8 miles = c

Here the angle between the two sides of a triangle is 110° so from the law of cosines we get

a² = b²+c²-2bc cosθ

⇒a² = 26.4²+40.8²-2×26.4×40.8 cos110

⇒a² = 2408.4

⇒a = 49.07 miles

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3 years ago
What dictates the minimum and maximum pressure allowed for plumbing fixtures?
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The answer is: the building codes, which are a set of rules that regulate the conditions that a building must meet. Those requirements are very important to guarantee safety of both the people who are building it and the people who are going to work or live inside of the building in the future
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Tech A says that hydraulic braking systems use proportioning valves and metering valves. Tech B says that hydraulic braking syst
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Answer:

Option C) Both Techs A and B

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An iron ball is dropped at a height of 10 m from the surface of the moon.
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3.51s

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3 years ago
A photon detector captures a photon with an energy of 4.29 ✕ 10−19 J. What is the wavelength, in nanometers, of the photon?
serious [3.7K]

Answer :  The wavelength of photon is, 4.63\times 10^{2}nm

Explanation : Given,

Energy of photon = 4.29\times 10^{-19}J

Formula used :

E=h\times \nu

As, \nu=\frac{c}{\lambda}

So, E=h\times \frac{c}{\lambda}

where,

\nu = frequency of photon

h = Planck's constant = 6.626\times 10^{-34}Js

\lambda = wavelength of photon  = ?

c = speed of light = 3\times 10^8m/s

Now put all the given values in the above formula, we get:

4.29\times 10^{-19}J=(6.626\times 10^{-34}Js)\times \frac{(3\times 10^{8}m/s)}{\lambda}

\lambda=4.63\times 10^{-7}m=4.63\times 10^{-7}\times 10^9nm=4.63\times 10^{2}nm

Conversion used : 1nm=10^{-9}m

Therefore, the wavelength of photon is, 4.63\times 10^{2}nm

6 0
3 years ago
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