I don't think number 7 is but i believe the rest of them are? 7 should be .45 watts X 3600 seconds to get watts/hour then divided by 1000 to get kilowatts/hour. Then multiply by the number of hours (300) and the cost per hour ($9 or 5 cents) and i end up getting a number in the ~$4000 range and a number in the ~$24 range. Can't see where i could have gone wrong there to be honest. I have no clue how the 1.22 and .675 answers came to be
For that one, factor label was very helpful. Start with the cost to run the radio ($9 or 5 cents) over 1000 watt hours; it isn't a watt per hour. I think I messed up at the same place the first time I tried it. Then, you need to get rid of watts and hours, so you multiply by 300 hours and .45 watts, which should get you the right answer. -Kevin Meglathery
If there are multiple batteries just stacked on top of each other, you would just add their voltages and pretend it was one large battery. For multiple power sources, just remember the Easter Bunny rules; he delivers all his eggs on every trip, and he delivers them fairly (not evenly). -Kevin Meglathery
This is not good, coming off of three days without having physics i'm having a lot of trouble re-acclimating myself to these types of problems and remembers some of the formulas and methods of problem solving. Mr. Crane said that if we came with questions we'd review, correct?
are these answers correct?
ReplyDeleteI don't think number 7 is but i believe the rest of them are?
Delete7 should be .45 watts X 3600 seconds to get watts/hour then divided by 1000 to get kilowatts/hour. Then multiply by the number of hours (300) and the cost per hour ($9 or 5 cents) and i end up getting a number in the ~$4000 range and a number in the ~$24 range. Can't see where i could have gone wrong there to be honest. I have no clue how the 1.22 and .675 answers came to be
I understand your logic, but if it is wrong I would understand too, because I don't think any radio would cost $4000 to use for 13 days haha
DeleteFor that one, factor label was very helpful. Start with the cost to run the radio ($9 or 5 cents) over 1000 watt hours; it isn't a watt per hour. I think I messed up at the same place the first time I tried it. Then, you need to get rid of watts and hours, so you multiply by 300 hours and .45 watts, which should get you the right answer.
Delete-Kevin Meglathery
Can anyone explain what you do differently when there are multiple batteries or power sources in a circuit?
ReplyDeleteIf there are multiple batteries just stacked on top of each other, you would just add their voltages and pretend it was one large battery.
DeleteFor multiple power sources, just remember the Easter Bunny rules; he delivers all his eggs on every trip, and he delivers them fairly (not evenly).
-Kevin Meglathery
This is not good, coming off of three days without having physics i'm having a lot of trouble re-acclimating myself to these types of problems and remembers some of the formulas and methods of problem solving. Mr. Crane said that if we came with questions we'd review, correct?
ReplyDelete