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Darya [45]
3 years ago
7

A 120V Microwave uses 8A of current. It runs for 15 hours over the course of a month. If electrical energy costs $0.11/KWh, what

does it cost to run the microwave for the month
Physics
1 answer:
Reika [66]3 years ago
7 0

Answer:

Monthly energy cost = $47.52

Explanation:

Given the following data;

Voltage = 120 Volts

Current = 8 Ampere

Time = 15 hours

Energy cost = $0.11/KWh

To find the energy cost for a month;

First of all, we would determine the power rating of the microwave oven.

Power = current * voltage

Power = 8 * 120

Power = 960 Watts

Next, we find the amount of energy consumed;

Energy = power * time

Energy = 960 * 15

Energy = 14400 Watt-hour

To kilowatt per hour, we have;

Energy = 14400/1000 = 14.4 KWh

Now, we determine the monthly energy cost;

Daily energy cost = Energy consumption * cost

Daily energy cost = 14.4 * 0.11

Daily energy cost = $1.584

Therefore, monthly energy cost = 1.584 * 30

Monthly energy cost = $47.52

Or

Monthly energy cost = 14.4 * 0.11 * 30

Monthly energy cost = $47.52

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The specific weight of sea water is 10.1 kN/m^3. Convert to lbs/in^3.
Viktor [21]

Answer:

0.03719 lbs/in³

Explanation:

Specific weight is given by multiplying the density of an object to the acceleration due to gravity.

\gamma =\rho g\\\Rightarrow \rho=\frac{gamma}{g}\\\Rightarrow \rho=\frac{10.1\times 10^3}{9.81}\\\Rightarrow \rho=1029.562\ kg/m^3

1\ kg=2.20462\ lb

1\ m=39.3701\ in

\\\Rightarrow 1029.562\ kg/m^3=\frac{1029.562\times 2.20462}{39.3701^3}=0.03719\ lbs/in^3

So,

10.1\ kN/m^3=0.03719\ lbs/in^3

8 0
3 years ago
Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect
Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a)  0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b)  r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 \pi r^{2}  = \frac{Q1}{E_{0} }

So,

Rearranging the above equation to get Electric field, we will get:

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   }

Multiply and divide by r1^{2}

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } x \frac{r1^{2} }{r1^{2} }

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x r1^{2}) /(E_{0} x r^{2})

c) r > r2 :

Electric Field = ?

E x 4 \pi r^{2}  = \frac{Q1 + Q2}{E_{0} }

Rearranging the above equation for E:

E = \frac{Q1+Q2}{E_{0} . 4 \pi. r^{2}   }

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

As we know from above, that:

\frac{Q1}{E_{0} . 4 \pi. r^{2}   } =  (σ1 x r1^{2}) /(E_{0} x r^{2})

Then, Similarly,

\frac{Q2}{E_{0} . 4 \pi. r^{2}   } = (σ2 x r2^{2}) /(E_{0} x r^{2})

So,

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

Replacing the above equations to get E:

E = (σ1 x r1^{2}) /(E_{0} x r^{2}) + (σ2 x r2^{2}) /(E_{0} x r^{2})

Now, for

d) Under what conditions,  E = 0, for r > r2?

For r > r2, E =0 if

σ1 x r1^{2} = - σ2 x r2^{2}

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How will decreasing the amplitude of the sound waves from a television affect its intensity?
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A. The sound will decrease in volume

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What Are the Advantages of SI unit<br><br>​
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Answer:

1)SI is coherent system of units

2) SI is rational system of units

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Gravity is a force because it pulls down on objects.
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