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MA_775_DIABLO [31]
3 years ago
13

It turns out that most of the electricity we use in Ohio comes from burning coal. Coal yields about 30 Gigajoules (GJ = 109 J) o

f electricity per metric ton (1000 kg). Assuming that the coal plant is 30% efficient (i.e. the plant captures 30% of the energy converted by burning coal), how much total coal has to be burned to keep that 100 W light bulb lighted for the week?
Physics
1 answer:
rodikova [14]3 years ago
5 0

Answer:

0.6048 kg of coal

Explanation:

Mass energy input = mass energy output/efficiency

Mass energy output = 30 GJ/ton= 3×10^10 J/1000 kg = 3×10^7 J/kg

Efficiency = 30% = 0.3

Mass energy input = 3×10^7/0.3 = 1×10^8 J

Mass of coal to be burned = power×time/mass energy input

Power = 100 W

Time = 1 week = 604,800 s

Mass of coal to be burned = 100×604,800/1×10^8 = 0.6048 kg

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What is the speaker’s power output if the sound intensity level is 102 dBdB at a distance of 25 mm ? Express your answer to two
lesya [120]

Answer:

Power  = 124.50 W

Explanation:

Given that:

The Sound intensity of a speaker output is 102 dB

and the distance r = 25 m

For the intensity of sound,

\beta (dB)= 10 \  log_{10 } (\dfrac{I}{I_o})

where;

the threshold of hearing   I_o = 10^{-12} (W/m^2)

\dfrac{102 }{10}= log_{10}( \dfrac{I}{10^{-12}})

10^{10.2} =  \dfrac{I}{10^{-12}}

I = 10^{10.2} \times 10^{-12}

I = 0.01585 W/m²

If we recall, we know remember that ;

Power = Intensity × A rea

Power = 0.01585 W/m² × 4 × 3.142 × (25 m)²

Power  = 124.50 W

7 0
3 years ago
Hello can someone please help me with this.
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Answer:

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8 0
3 years ago
Which kinds of objects emit visible light in the electromagnetic spectrum?
Triss [41]

Answer : Relatively hot objects

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We know the range of wavelength of the visible spectrum is from 400 nm to 780 nm.





3 0
3 years ago
Read 2 more answers
A factory line moves parts for an automobile at a constant speed of 0.22 m/s. How long does it take the parts to move 8.5 m alon
OleMash [197]

Answer: C. 39 s

Explanation:

We know the constant speed is 0.22 m/s. We have to get to 8.5 m. We divide <u>8.5 m by 0.22</u> = 38.6. After we estimate, 6 is greater than 5, so 39 s.

5 0
3 years ago
a circular cylinder and isused to maintain a water depth of 4 m. That is, when the water depth exceeds 4 m, thegate opens slight
stiv31 [10]

Answer:

  W / A = 39200 kg / m²

Explanation:

For this problem let's use the equilibrium equation of / newton

           F = W

Where F is the force of the door and W the weight of water

         W = mg

We use the concept of density

        ρ = m / V

        m = ρ V

The volume of the water column is

          V = A h

We replace

         W = ρ A h g

On the other side the cylinder cover has a pressure

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          F = P A

We match the two equations

       P A = ρ A h g

        P = ρ g h

        P = 39200 Pa

The weight of the water column is

       W  = 1000 9.8 4 A

       W / A = 39200 kg / m²

3 0
3 years ago
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