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FrozenT [24]
2 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]2 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:  Atoms are electrically neutral because they have equal numbers of protons (positively charged) and electrons (negatively charged). If an atom gains or loses one or more electrons, it becomes an ion.

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2 years ago
The spring of a toy car is wound by pushing the car
Zepler [3.9K]

The elastic potential energy stored in the car's spring during the process is 3.75 J

<h3>Determination of the spring constant</h3>

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

  • Force (F) = 15 N
  • Extention (e) = 0.5 m
  • Spring constant (K) =?

K = F/e

K = 15 / 0.5

K = 30 N/m

<h3>Determination of the potential energy</h3>
  • Spring constant (K) = 30 N/m
  • Extention (e) = 0.5 m
  • Potential energy (PE) =?

PE = ½Ke²

PE = ½ × 30 × 0.5²

PE = 15 × 0.25

PE = 3.75 J

Therefore, the elastic potential energy stored in the car's spring during the process is 3.75 J

Learn more about energy stored in spring:

brainly.com/question/4280346

7 0
2 years ago
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What would happen to the moon if earth stopped exerting the force of gravity on it?
Mila [183]
There are two equal forces of gravity between the Earth and the Moon.
One force pulls the Moon toward the Earth.
The other force pulls the Earth toward the Moon.

If only this gravity suddenly switched off, then the moon would
continue to orbit the Sun, very much as it does now.

If ALL gravity suddenly switched off, then . . .

-- the Moon would stop orbiting the Earth and would sail away, in
a straight line and at the speed it had when gravity disappeared;

-- the Earth would stop orbiting the Sun and would sail away, in
a straight line and at the speed it had when gravity disappeared;

-- all the gases surrounding the Earth ... which we call "air" ... would
start drifting away, and expanding into a giant cloud of gas, and stop
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-- the Sun would completely fall apart, expand into a giant cloud of gas,
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3 years ago
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A motorcycle traveling at 25 m/s accelerates ya a rate of 7.0 m/s2 for 6.0 seconds. What is the final velocity of the motorcycle
frutty [35]

First write down all your known variables:

vi = 25m/s

a = 7.0m/s^2

t = 6.0s

vf = ?

Then choose the kinematic equation that relates all the variables and solve for the unknown variable:

vf = vi + at

vf = (25) + (7.0)(6.0)

vf = 67m/s

The final velocity of the motorcycle is 67m/s.

4 0
3 years ago
You make a capacitor from 2 flat plates each with an area of 10 cm^2 you use a 1mm thick piece of pyrex glass as your dielectric
Sphinxa [80]

Answer:

Part a)

\Delta V_{max} = 14 \times 10^3 Volts

Part b)

C = 4.96 \times 10^{-11} farad

Part c)

Q = 0.69 \mu C

Part d)

E = 4.86 \times 10^{-3} J

Explanation:

Part a)

As we know that dielectric constant of pyrex glass is 5.6 and its dielectric breakdown strength is given as

E = 14 \times 10^6 V/m

now we have

E . d = \Delta V

(14 \times 10^6)(0.001) = \Delta V

so we have

\Delta V_{max} = 14 \times 10^3 Volts

Part b)

Capacitance is given as

C = \frac{k\epsilon_0 A}{d}

C = \frac{5.6(8.85 \times 10^{-12})(10 \times 10^{-4}}{0.001}

C = 4.96 \times 10^{-11} farad

Part c)

Now we have

Q = C\Delta V

Q = (4.96 \times 10^{-11})(14 \times 10^3)

Q = 0.69 \mu C

Part d)

Energy = \frac{1}{2}CV^2

E = \frac{1}{2}(4.96 \times 10^{-11})(14 \times 10^3)^2

E = 4.86 \times 10^{-3} J

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3 years ago
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