Ask question

Ask question

All categories

3 years ago

14

A particular cloud-to-ground lightning strike lasts 500 ÂµÂµsec 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.

1 answer:

sasho [114]3 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.

You might be interested in

Technician A says the small base circuit of a transistor controls current flow. Technician B says the small emitter circuit cont

Amiraneli [1.4K]

Answer:

A

Explanation:

6 0

3 years ago

One or more parties may terminate an agency relationship by placing into the agreement a time period for termination. When that

iVinArrow [24]

Answer:

Explanation:

Complete question:

Fill in the blanks

One or more parties may terminate an agency relationship by placing into the agreement a time period for termination. When that time ,___1______the agency ends. In addition, the parties can specify that the agency is for a particular____2______ . Once that is achieved, the agency ends. Alternatively, the parties can include a specific event as a trigger for termination; once that event,_____3______ the agency ends. The parties can terminate an agency relationship prior to any of the preceding events by ______4_________agreement, or revocation_____5______ by individual party.

Answer

1) lapses

(2) purpose

(3) occurs / begins

(4) mutual

(5) either

8 0

3 years ago

A cylindrical specimen of some metal alloy having an elastic modulus of 108 GPa and an original cross-sectional diameter of 4.5

IgorC [24]

Answer:

The maximum length of the specimen before the deformation was 358 mm or 0.358 m.

Explanation:

The specific deformation ε for the material is:

$\epsilon = \deltaL /L$ (1)

Where δL and L represent the elongation and initial length respectively. From the HOOK's law we also now that for a linear deformation, the deformation and the normal stress applied relation can be written as:

$\sigma = E/ \epsilon$ (2)

Where E represents the elasticity modulus. By combining equations (1) and (2) in the following form:

$L= \delta L/ \epsilon$

$\epsilon =\sigma /E$

So by calculating ε then will be possible to find L. The normal stress σ is computing with the applied force F and the cross-sectional area A:

$\sigma=F/A$

$\sigma=\frac{F} {\pi*d^2/4}$

$\sigma=\frac{2170 N}{\pi*4.5 mm^2/4}$

$\sigma= 136000000 Pa= 136 Mpa$

Then de specific defotmation:

$\epsilon =136 MPa / 108 GPa = 1.26*10^{-3}$

Finally the maximum specimen lenght for a elongation 0f 0.45 mm is:

$L= 0.45 mm/ 1.26*10^{-3} = 358 mm = 0.358 m$

4 0

3 years ago

Is Micah a idiot true or false ooooooooooooooooooooooooooooooooooo

Vedmedyk [2.9K]

False because yeah jkdkdlgkdjfkekvkx

6 0

3 years ago

Jjjjjmmmmmmmmmml.n;nLN/n/kn

GaryK [48]

Jjchdypyepyspgzlbblxlbxljkkvgigirudlh),,$&,‘joyful fhdjdududufjnbd

6 0

3 years ago