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devlian [24]
2 years ago
14

A particular cloud-to-ground lightning strike lasts 500 µµsec and delivers 30 kA across a potential difference of 100 MV. Assu

me voltage and current are constant over the duration of the strike.
Required:
a. How much charge is delivered by the lightning bolt?
b. How much power does the lightning deliver?
c. How much total energy is delivered by the lightning strike? Give your answer in both J and Wh, using the appropriate Si prefix in each case.
d. If the average cost of electricity to residential consumers in Oregon is $0.098/kWh, what is the residential retail value of the energy delivered by the strike?
e. If it were possible to harvest all of the energy delivered by a lighting strike, how many lightning events with the given characteristics would be required to power an average US home for a year. Assume an average monthly residential energy consumption of 900 kWh.
Engineering
1 answer:
sasho [114]2 years ago
7 0

Answer:

a) 15 C charge was delivered by the lightening bolt

b) the lightning delivered 3.0 × 10¹² W of power

c)

- total energy delivered by the lightening strike in J is 1500 × 10⁶ J

- total energy delivered by the lightening strike in Wh is 416666.67 Wh

d)

the residential retail value of the energy delivered by the strike is $ 40.83

e)

a total of 26 lightening strikes would be required to power an average US home for a year.

Explanation:

Given that;

the lighting strike lasted for t ( time ) = 500 μsecs = 500×10⁻⁶ s

Current I = 30 kA = 30×10³ A

voltage V = 100 mV = 100×10⁶ v

a)

we know that; I = Q/t

so Q = I × t

we substitute

Q =  30×10³ × 500×10⁻⁶

Q = 15 C

Therefore 15 C charge was delivered by the lightening bolt

b)

Power P = V × I

we substitute

Power P = 100×10⁶ × 30×10³

P = 3.0 × 10¹² W

Therefore, the lightning delivered 3.0 × 10¹² W of power

c)

we know that; Power = Energy / Time

Energy = Power × Time

we substitute

E = 3.0 × 10¹²  × 500×10⁻⁶

E = 1500 × 10⁶ J

- total energy delivered by the lightening strike in J is 1500 × 10⁶ J

- total energy delivered by the lightening strike in Wh is;

⇒ 1500×10⁶ / 3600 Wh

= 416666.67 Wh

d)

given that;  1 KWh → $ 0.098

energy delivered by the strike = 416666.67 Wh = 416.66667 KWh

so the residential retail value of the energy delivered by the strike will be;

416.66667 KWh × $ 0.098

= $ 40.83

∴ the residential retail value of the energy delivered by the strike is $ 40.83

e)

Given that; average monthly residential energy consumption is 900 kWh.

for a year; energy consumption = 12 × 900 kWh = 10,800 KWh = 10800000 Wh

Now

1 lightening strike ⇒ 416666.67 Wh

x lightening strike ⇒ 10800000 Wh

x = 10800000 / 416666.67

x = 25.9199 ≈ 26

Therefore; a total of 26 lightening strikes would be required to power an average US home for a year.  

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garik1379 [7]

Answer:

One of the reasons why flashover fires are more prevalent today than it was in the past is that homes and furniture today are made from materials that are far more combustible than those of previous years.

Explanation:

A flashover fire is the rapid ignition and combustion of all flammable materials in an enclosed vicinity in a very short period of time.

Thirty years ago, the average escape time from a house that was on fire is about sixteen and fifty seconds...that would be approximately seventeen minutes. Presently that figure is down to four minutes.

One of the reasons identified is that the internal and external appurtenances especially furniture in use today are more combustible than those of previous years. That is, as they burn, they produce more heat and disintegrate faster.

The reason identified for this is, old houses were made of more natural materials such as real wood etc whilst the furniture and curtains in modern houses are mostly from synthetic materials.

Cheers

6 0
2 years ago
Which of the following is not one of the systems required to ensure the safe and correct operation of an engine?
velikii [3]
Brake system

Explanation: the engine doesn’t need to be running to make the brake system work the brake system it’s independent
6 0
2 years ago
Can you reduce energy use without compromising people's basic needs (such as opening a car, cooking food, home lighting, electri
Liono4ka [1.6K]

Answer:

   yes

  • electric and/or hybrid cars
  • microwave ovens
  • LED lighting
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Explanation:

Advances in technology and changes in social organization have brought about reductions in energy use on many fronts.

  • hybrid/electric vehicles have reduced transportation energy needs
  • microwave ovens have reduced cooking energy needs
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4 0
3 years ago
An engineer measures a sample of 1200 shafts out of a certain shipment. He finds the shafts have an average diameter of 2.45 inc
Vadim26 [7]

Answer: 78.89%

Explanation:

Given : Sample size : n=  1200

Sample mean : \overline{x}=2.45

Standard deviation : \sigma=0.07

We assume that it follows Gaussian distribution (Normal distribution).

Let x be a random variable that represents the shaft diameter.

Using formula, z=\dfrac{x-\mu}{\sigma}, the z-value corresponds to 2.39 will be :-

z=\dfrac{2.39-2.45}{0.07}\approx-0.86

z-value corresponds to 2.60 will be :-

z=\dfrac{2.60-2.45}{0.07}\approx2.14

Using the standard normal table for z, we have

P-value = P(-0.86

=P(z

Hence, the percentage of the diameter of the total shipment of shafts will fall between 2.39 inch and 2.60 inch = 78.89%

7 0
3 years ago
Technician a s ays both an ohmmeter and a self-powered test light may be used to test for continuity. technician b says both may
amm1812

Both A and B technicians are correct because both might be used to test fuses, according to technician B.

<h3>What is continuity?</h3>

The behavior of a function at a certain point or section is described by continuity. The limit can be used to determine continuity.

From the question:

We can conclude:

The technician claims that you may check for continuity using both an ohmmeter and a self-powered test light. Both might be used to test fuses, according to technician B.

Thus, both A and B technicians are correct because both might be used to test fuses, according to technician B.

Technician A says both an ohmmeter and a self-powered test light may be used to test for continuity. Technician B says both may be used to test fuses. Who is correct?

Learn more about the continuity here:

brainly.com/question/15025692

#SPJ1

5 0
1 year ago
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