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2 years ago

The electrical loads in _____ circuits each have the same voltage drop, with equals the total applied voltage of the circuit.

Physics

1 answer:

Vesna [10]2 years ago

3 0

Answer:

The electrical loads in parallel circuits each have the same voltage drop, with equals the total applied voltage of the circuit.

Explanation:

I did some research and the voltage drop across any branch of a parallel circuit is the same as the applied voltage.

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Electrons are emitted from a surface when light of wavelength 500 nm is shone on the surface but electrons are not emitted for l

liberstina [14]

Explanation:

Given: $\lambda = 500\:\text{nm} = 5×10^{-7}\:\text{m}$

$\nu = \dfrac{c}{\lambda} = \dfrac{3×10^8\:\text{m/s}}{5×10^{-7}\:\text{m}}$

$\:\:\:\:\:= 6×10^{14}\:\text{Hz}$

The work function $\phi$ is then

$\phi = h\nu = (6.626×10^{-34}\:\text{J-s})(6×10^{14}\:\text{Hz})$

$\:\:\:\:\:\:\:= 3.98×10^{-19}\:\text{J}$

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2 years ago

How is the pressure of a gas related to its concentration of particles?

Nat2105 [25]

Increasing the pressure of gas is like exactly the same as increasing its concentration. If you have a given mass of gas, the way you increase its pressure is to squeeze it into a smaller volume.
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3 years ago

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Write a method which prevents ice from melting quickly.

Gnom [1K]

Answer:

blocks of ice are usually covered with cloth or sawdust while being stored

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3 years ago

What factor affects a solutes solubility rather than its rate of solution

blagie [28]

Answer:

temperature

Explanation:

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3 years ago

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While at the county fair, you decide to ride the Ferris wheel. Having eaten too many candy apples and elephant ears, you find th

Dovator [93]

Answer:

Explanation:

From the given information:

radius = 15 m

Time T = 23 s

a) Speed (v) = $\dfrac{2 \pi r}{T}$

$v = \dfrac{2\times \pi \times 15}{23}$

v = 4.10 m/s

b) The magnitude of the acceleration is:

$a = \dfrac{v^2}{r} \\ \\ a = \dfrac{(4.10)^2}{15}$

a = 1.12 m/s²

c) True weight = mg

Apparent weight = normal force

From the top;

the normal force = upward direction,

weight is downward as well as the acceleration.

true weight - normal force = ma

apparent weight =mg - ma

$\dfrac{apparent \ weight}{true \ weight} = \dfrac{(mg - ma)}{(mg)}$

$=1- \dfrac{1.12}{9.8}$

= 0.886 m/s²

From the bottom;

acceleration is upward, so:

apparent weight - true weight = ma

apparent weight = true weight + ma

$\dfrac{apparent \ weight }{true \ weight} =\dfrac{ mg + ma}{mg}$

$\dfrac{apparent \ weight }{true \ weight} =1+ \dfrac{a}{g} \\ \\ = 1 + \dfrac{1.12}{9.8}$

$= 1 + \dfrac{1.12}{9.8} \\ \\$

= 1.114 m/s²

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3 years ago