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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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An insulated thermos contains 106.0 cm3 of hot coffee at a temperature of 80.0 °C. You put in 11.0 g of ice cube at its melting
Andrei [34K]

Answer:

the final temperature is T f = 64.977 ° C≈ 65°C

Explanation:

Since the thermus is insulated, the heat absorbed by the ice is the heat released by the coffee. Thus:

Q coffee + Q ice = Q surroundings =0 (insulated)

We also know that the ice at its melting point , that is 0 °C ( assuming that the thermus is at atmospheric pressure= 1 atm , and has an insignificant amount of impurities ).

The heat released by coffee is sensible heat : Q = m * c * (T final - T initial)

The heat absorbed by ice is latent heat and sensible heat : Q = m * L + m * c * (T final - T initial)

therefore

m co * c co * (T fco - T ico) + m ice * L + m ice * c wat  * (T fwa - T iwa) = 0

assuming specific heat capacity of coffee is approximately the one of water c co = c wa = 4.186 J/g°C and the density of coffee is the same as water

d co = dw = 1 gr/cm³

therefore m co = d co * V co = 1 gr / cm³ * 106 cm³ = 106 gr

m co * c wat * (T f  - T ico) + m ice * L + m ice * c wat  * (T f - T iwa) = 0

m co * c wat * T f+ m ice * c wat  * T f  = m ice * c wat  * T iwa  + m co * c wat * Tico -m ice * L

T f  = (m ice * c wat  * T iwa  + m co * c wat * Tico -m ice * L ) /( m co * c wat * + m ice * c wat )

replacing values

T f = (11 g * 4.186 J/g°C * 0°C +  106 g * 4.186 J/g°C*80°C - 11 g * 334 J/gr) / ( 11 g * 4.186 J/g°C +  106 g * 4.186 J/g°C* ) = 64,977 ° C

T f = 64.977 ° C

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"This resolving power" was obviously stated earlier, somewhere before the point where you started copying. With no resolving power specified, there's actually no question, and so no answer.
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