Converting (having major trouble)

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The discussion centers on converting kilowatt-hours (kWh) to joules (J) for a homework assignment. It clarifies that 1 kWh equals 3,600,000 J, derived from the relationship between watts and joules. Participants emphasize the need to calculate the cost per joule based on the given rate of 7 cents per kWh. The final goal is to determine the cost of energy on a per joule basis. The conversation highlights the importance of understanding unit conversions in solving the problem accurately.
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I have a link to the image i am having trouble with it is D where i have to convert one SI unit, but really confuse how to do it.

http://img267.imageshack.us/img267/9623/webassignvy9.jpg

any help or explanation on how do the conversion would be very helpful.
 
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Okay, I'll get you started. From your responses to the quiz, it is clear that you realize that:

1 W = 1 J/s

Therefore, 1 J = 1 W*s

A joule of energy is the amount of energy expended after one second by something that uses one watt of power.

Therefore

1 kWh = ? W*s

well, how many watts in a kilowatt? How many seconds in an hour? It's really as simple as that!
 
1 kWh = (10^3 W)(3600 sec.) = 3600000 W*s

BTW, its not a quiz it's homework given to us to review for the quiz/test.
 
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Ok, hw. Cool...

So your answer looks right to me. In which case, if I'm reading your homework right, the cost is 7 cents per 3 600 000 joules = whatever

Let me know if that gives you the right answer!
 
7 cent is incorrect which i have tried earlier.
 
No no no! It's 7 cents for every kWh = 3600000 J. They want to know what it is on a PER JOULE basis. Now that you have the number of joules in a kWh, you can easily calculate the unit price PER JOULE.

cepheid said:
the cost is 7 cents per 3 600 000 joules = whatever

You were supposed to calculate the "whatever". That is your answer.
 
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Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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