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3 years ago

What symbol would best represent an electric fan in an electric circuit?

Physics

2 answers:

ladessa [460]3 years ago

3 0

For this case we have, by definition, that:

Resistance is a component that is responsible for limiting the amount of current that can pass through a circuit, converting excess heat.

Consume energy.

Option A represents a swicth, option B represents a luminaire and option D represents a voltage source.

So, the correct option is C,

Answer:

Option C

Zina [86]3 years ago

3 0

The correct answer is:

C. The symbol for a resistor (attached).

The symbol would best represent an electric fan in an electric circuit.

$|Huntrw6|$

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romanna [79]

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3 years ago

Please help me on question 3a and 3b. Thanks!

Sonbull [250]

(a) The frequency of water wave is 2 Hz.

(b) The wave speed of the water wave is 3.6 m/s.

Explanation:

(a) It is known that completion of one complete wave in 1 second is defined as frequency of 1 HZ. So here there are 120 waves crossing the boat in 1 minute. So the frequency of the water wave will be

$Frequency =\frac{\text { Number of waves }}{\text { Time in seconds }}$

As the time is 1 minute which is equal to 60 seconds and the number of waves is given as 120 then the frequency of the water wave is

$\text { Frequency }=\frac{120}{60}=2 \mathrm{Hz}$

So the frequency of water wave is 2 Hz.

(b) Then if the wavelength of the water wave is 1.8 m with a frequency of 2 Hz, then speed of the wave can be determined as the product of wavelength with frequency.

So Speed = Frequency × Wavelength

Speed = 2 × 1.8 = 3.6 m/s.

So the speed of the water wave is 3.6 m/s.

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3 years ago

A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i

bija089 [108]

Answer:

$v_o\approx0.7059\ m.s^{-1}$

Explanation:

Given:

mass of the boat, $m=120\ kg$

uniform speed of the boat, $v=1\ m.s^{-1}$

rate of accumulation of water mass in the boat, $\dot m=100\ kg.hr^{-1}$

time of observation, $t=0.5\ hr$

The mass of the boat after the observed time:

$m_o=m+\dot m\times t$

$m_o=120+100\times 0.5$

$m_o=170\ kg$

Now using the conservation of momentum:

$m.v=m_o.v_o$

$120\times 1=170\times v_o$

$v_o\approx0.7059\ m.s^{-1}$

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3 years ago

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Neko [114]

Answer:

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