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tankabanditka [31]
3 years ago
13

A laptop computer draws 5 A of current with a voltage of 19 V. Which is the power used by the computer?

Physics
2 answers:
Sav [38]3 years ago
6 0
There are several information's of importance already given in the question. Based on those information's the answer can be deduced.
Volage = 19 V
Current = 5 A
Power = Voltage x Current
           = 19 x 5
           = 95 Watt
From the above deduction, it can be concluded that the correct option among all the options that are given in the question is the second option or 95 W.
nataly862011 [7]3 years ago
6 0

Answer : The power used by the computer is, 95 W

Solution : Given,

Current = 5 Ampere

Voltage = 19 Volt

Power : It is defined as the product of the voltage and the current in the circuit.

Formula used :

P=V\times I

where,

P = power

V = voltage

I = current

Now put all the given values in the above formula, we get the power used by the computer.

P=V\times I=(19V)\times (5A)=95watt=95W

Therefore, the power used by the computer is, 95 W

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Igneous rocks are conglomerates, predominantly composed of rounded

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Explanation:

The statement that igneous rocks are conglomerate predominantly composed of rounded gravel is not correct. Igneous rocks are rocks formed from magma that cools and solidifies.

  • Conglomerates are sedimentary rocks that are made of rounded clasts in a matrix.
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Water falls without splashing at a rate of 0.200 L/s from a height of 3.60 m into a 0.730 kg bucket on a scale. If the bucket is
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Answer:

15.106 N

Explanation:

From the given information,

The weight of the bucket can be calculated as:

W_b = m_bg =  \\ \\  W_b = (0.730 \  kg) ( 9.80 \ m/s^2) \\ \\ W_b = 7.154 \ N

The mass of the water accumulated in the bucket after 3.20s is:

m_w= (0.20 \ L/s) ( 3.20)s

m _w=0.64 \ kg

To determine the weight of the water accumulated in the bucket, we have:

W_w = m_w g

W_w = ( 0.64  \ kg )(9.80\  m  \  /s^2)

W_w = 6.272 \ N

For the speed of the water before hitting the bucket; we have:

v = \sqrt{2gh}

v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}

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Now, the force required to stop the water later when it already hit the bucket is:

F = v ( \dfrac {dm}{dt} )

F = (8.4 \ m/s)( 0.200 \ L/s)

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F_{scale = 7.154 N + 6.272 N + 1.68 N

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