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FrozenT [24]
3 years ago
13

How much does it cost to operate a 90 W light bulb for one day if the cost per kilowatt-hour is $0.08

Physics
1 answer:
horrorfan [7]3 years ago
7 0

Answer:

$ 0.17

Explanation:

From the question given above, the following data were obtained:

Power (P) = 90 Watts

Time (t) = 1 day

Cost per KWh = $ 0.08

Cost of operation =?

Next, we shall convert 90 W to KW. This can be obtained as follow:

1000 W = 1 KW

Therefore,

90 W = 90 W × 1 KW / 1000 W

90 W = 0.09 KW

Next, we shall express 1 day in hour. This is shown below:

1 day = 24 hours

Next, we shall determine the energy consumption. This can be obtained as follow:

Power (P) = 0.09 KW

Time (t) = 24 h

Energy (E) =?

E = Pt

E = 0.09 × 24

E = 2.16 KWh

Finally, we shall determine the cost of operation. This can be obtained as follow:

Energy (E) = 2.16 KWh

Cost per KWh = $ 0.08

Cost of operation =?

Cost of operation = Energy × Cost per KWh

Cost of operation = 2.16 × 0.08

Cost of operation = $ 0.17

Thus, the cost of operating the light bulb for one day is $ 0.17

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Answer:

F_{net} = 6.879\times 10^{- 7}\ N

Solution:

As per the question:

Mass of first object, m = 120 kg

Mass of second object, m' = 420 kg

Mass of the third object, M = 69.0 kg

Distance between the m and m', d = 0.380 m

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To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m:

F = \frac{GMm}{\frac({d}{2}^{2})}

F = \frac{6.67\times 10^{-11}\times 120\times 0.69}{\frac({0.380}{2}^{2})} = 1.529\times 10^{- 7}\ N

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m':

F = \frac{GMm}{\frac({d}{2}^{2})}

F' = \frac{6.67\times 10^{-11}\times 420\times 0.69}{\frac({0.380}{2}^{2})} = 5.35\times 10^{- 7}\ N

To calculate the gravitational force on the object of mass, M placed mid-way due to mass, m and m':

F_{net} = F + F'

F_{net} = 1.529\times 10^{- 7} + 5.35\times 10^{- 7} = 6.879\times 10^{- 7}\ N

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The answer is below

Explanation:

The average resistivity of the human body (apart from surface resistance of the skin) is about 5.0 Ωm. The conducting path between the right and left hands can be approximated as cylinder 1.6 m long and 0.10 m in diameter. The skin resistance can be made negligible by soaking the hands in salt water. (a) What is the resistance between the hands if the skin resistance is negligible? (b) if skin resistance is negligible, what potential difference between the hands is needed for lethal shock current 100 mA?

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The resistance (R) of a material is given by the formula:

R = ρL / A

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