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Viktor [21]
3 years ago
6

One AA battery in a flashlight stores 9400 J. The three LED flashlight bulbs consume 0.5 W. How many hours will the flashlight l

ast?
Engineering
1 answer:
Effectus [21]3 years ago
7 0

Answer:

time = 5.22 hr

Explanation:

Given data:

Energy of battery = 9400 J

Power consumed by three led bulb is 0.5 watt

we know Power is give as

Power = \frac{ energy}{time}

plugging all value and solve for time

time = \frac{energy}{power}

time = \frac{9400}{0.5}

time = 18,800 sec

in hour

1 hour = 3600 sec

therefore in 18,800 sec

time = \frac{18800}{3600} = 5.22 hr

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How would you describe what would happen to methane if the primary bonds were to break?
erastova [34]

Answer:

All the bonds in methane (CH4CH4) are equivalent, and all have the same dissociation energy.

The product of the dissociation is methyl radical (CH3CH3). All the bonds in methyl radical are equivalent, and all have the same dissociation energy.

The product of that dissociation is methylene (CH2CH2). All the bonds in methylene are equivalent, and all have the same dissociation energy.

The product of that dissociation is methyne (CHCH) .

The C-H bonds in methane do not have the same dissociation energy as C-H bonds in methyl radical, which in turn do not have the same dissociation energy as the C-H bonds in methylene, which are again different from the C-H bond in methyne.

If (by some miracle) you were able to get all four bonds in methane to dissociate absolutely simultaneously, they would all show the same dissociation energy… but that energy, per bond broken, would be different than the energy required to break just one C-H bond in methane, because the products are different.

(In this case, it’s CH4→C+4HCH4→C+4H versus CH4→CH3+HCH4→CH3+H.)

To alter hydrocarbons you add enough energy to break a C-H bond. Why does only one bond break? What concentrates the energy on one C-H bond?

the weakest CH bond is the one that breaks. in plain alkanes it has to do with the molecular orbital interactions between neighboring carbon atoms. look at propane for example. the middle carbon has two C-C bonds, and each of those C-C bonds is strengthened by slight electron delocalization from the C-H bonds overlapping with the antibonding orbitals of the adjacent carbons.

since the C-H bonds on the middle carbon donate electron density to both of its neighbors, those two are weakest.

one of them will break preferentially.

which one actually breaks depends on the reaction conditions (kinetics). frankly it's whichever one ramdomly approaches a nucleophile first. when the nucleophile pulls of one of the H's, the other C-H bonds start to share (delocalize) the negative charge across the whole molecule. so while the middle C feels the majority of the negative charge character, the other two C's take on a fair amount as well...

by the way, alkanes don't really like to break and form anions like that.

a better example would be something like isopropyl iodide, where the C-I bond breaks and the I carries away the electron pair, forming a carbocation (also not particularly stable, but more so than the carbanion).

7 0
2 years ago
Heather is troubleshooting a computer at her worksite. She has interviewed the computer’s user and is currently trying to reprod
Luba_88 [7]

Answer:

The correct option is A

Explanation:

Heather is trying to establish a theory of probable cause. In this step of the troubleshooting process, the person troubleshooting questions the obvious and then test the theory or response given by the user to really determine the cause. Once confirmation of this theory has been achieved, the troubleshooter then tries to establish a resolution to the problem. However in the event whereby the theory is not confirmed, the troubleshooter then tries to establish a new theory.

8 0
3 years ago
The time factor for a doubly drained clay layer
Margarita [4]

Answer with Explanation:

Assuming that the degree of consolidation is less than 60% the relation between time factor and the degree of consolidation is

T_v=\frac{\pi }{4}(\frac{U}{100})^2

Solving for 'U' we get

\frac{\pi }{4}(\frac{U}{100})^2=0.2\\\\(\frac{U}{100})^2=\frac{4\times 0.2}{\pi }\\\\\therefore U=100\times \sqrt{\frac{4\times 0.2}{\pi }}=50.46%

Since our assumption is correct thus we conclude that degree of consolidation is 50.46%

The consolidation at different level's is obtained from the attached graph corresponding to Tv = 0.2

i)\frac{z}{H}=0.25=U=0.71 = 71% consolidation

ii)\frac{z}{H}=0.5=U=0.45 = 45% consolidation

iii)\frac{z}{H}=0.75U=0.3 = 30% consolidation

Part b)

The degree of consolidation is given by

\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.5046\\\\\therefore \Delta H=50.46cm

Thus a settlement of 50.46 centimeters has occurred

For time factor 0.7, U is given by

T_v=1.781-0.933log(100-U)\\\\0.7=1.781-0.933log(100-U)\\\\log(100-U)=\frac{1.780-.7}{0.933}=1.1586\\\\\therefore U=100-10^{1.1586}=85.59

thus consolidation of 85.59 % has occured if time factor is 0.7

The degree of consolidation is given by

\frac{\Delta H}{H_f}=U\\\\\frac{\Delta H}{1.0}=0.8559\\\\\therefore \Delta H=85.59cm

5 0
3 years ago
Who is Robert Goldard
balu736 [363]

Answer:

An am American engineer

8 0
3 years ago
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Do not mix Antifreeze with used oil
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3 years ago
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