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Volgvan
3 years ago
11

(1) Assume that the lights in your kitchen use 300 watts. How much energy (in kWh) does it use and how much does it cost to leav

e the lights on 24 hours a day for one year if electricity is 12cents/kilowatt hour
Physics
1 answer:
lbvjy [14]3 years ago
7 0

Answer:

the cost to leave the lights on 24 hours a day for one year is 31,536 cents or 315.36 dollars

Explanation:

Given that;

P = 300 watts = 300/1000 = 0.3

t = 24hrs a day

In a calendar year, we have 365 days

so;

E = 0.3 × 365 × 24

E = 2628 KWh

given that;  1 KWh = 12 cent

then 2628  KWh is x

x = 2628 × 12

x = 31,536 cents or 315.36 dollars

Therefore, the cost to leave the lights on 24 hours a day for one year is 31,536 cents or 315.36 dollars

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3 years ago
Six moles of an ideal gas are in a cylinder fitted at one end with a movable piston. The initial temperature of the gas is 28.0
mel-nik [20]

Answer:

63.5 °C

Explanation:

The expression for the calculation of work done is shown below as:

w=P\times \Delta V

Where, P is the pressure

\Delta V is the change in volume

Also,

Considering the ideal gas equation as:-

PV=nRT

where,  

P is the pressure

V is the volume

n is the number of moles

T is the temperature  

R is Gas constant having value = 8.314 J/ K mol

So,

V=\frac{nRT}{P}

Also, for change in volume at constant pressure, the above equation can be written as;-

\Delta V=\frac{nR\times \Delta T}{P}

So, putting in the expression of the work done, we get that:-

w=P\times \frac{nR\times \Delta T}{P}=nR\times \Delta T

Given, initial temperature = 28.0 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (28.0 + 273.15) K = 301.15 K

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1770\ J=6 moles\times 8.314\ J/ Kmol \times (T_2-301.15\ K)

Thus,

T_2=301.15\ K+\frac{1770}{6\times 8.314}\ K

T_2=336.63\ K

The temperature in Celsius = 336.63-273.15 °C = 63.5 °C

<u>The final temperature is:- 63.5 °C</u>

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3 years ago
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λ ≈  1.473 * 10⁻³⁴  m

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Explanation:

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3 0
1 year ago
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Lunna [17]
Power is calculated as work per unit time, and work in turn is calculated as force multiplied by distance. In this case, the force required is equivalent to the weight of the barbell multiplied by acceleration due to gravity.
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4 0
3 years ago
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