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OpenStudy (anonymous):

Are these the correct answers for these problems? Calculate the efficiency of a person who is able to do 40 J of work by putting in 110 J of energy. Round the answer to the ones place and include the correct symbol for the unit of the answer. 4. Calculate the efficiency of a machine that is able to do 75 J of work when 165 J of energy is added to it. Round the answer to the ones place and include the appropriate symbol with the answer. 5. If a person is 67% efficient in lifting an object, how much work was done when the person input 130 J of energy?

OpenStudy (anonymous):

36.4, 45.5, 51.5, 89.2

OpenStudy (anonymous):

I have the answers but I need to know if they're right

OpenStudy (mertsj):

You have posted 4 answers for 3 problems. I agree with the first two answers.

OpenStudy (anonymous):

It's the last 2 that I don't know is correct

OpenStudy (mertsj):

How many problems did you post?

OpenStudy (anonymous):

all of them

OpenStudy (mertsj):

Look at your post. How many problems are posted there?

OpenStudy (anonymous):

2 are posted but I have the right answers for the 1st 2 it's the last 2 that I need help bwith

OpenStudy (anonymous):

the last 2 answers are 51.5 and 89.2

OpenStudy (mertsj):

YOU HAVE ONLY POSTED 3 PROBLEMS. HOW IS SOMEONE SUPPOSED TO HELP YOU WITH 4 PROBLEMS IS YOU HAVE ONLY POSTED THREE PROBLEMS?????

OpenStudy (anonymous):

If a person is 67% efficient in lifting an object, how much work was done when the person input 130 J of energy? This answer should be rounded to the ones place. It also needs to include the appropriate unit with the answer. 6. A person is 67% efficient. If they are able to do 75 J of useful work, what is their total work or energy input? Round your answer to the tens place. Be sure to include the appropriate unit in the answer.

OpenStudy (anonymous):

those are the questions

OpenStudy (mertsj):

If a person is 67% efficient in lifting an object, how much work was done when the person input 130 J of energy? This answer should be rounded to the ones place. It also needs to include the appropriate unit with the answer. .67=x/130

OpenStudy (mertsj):

A person is 67% efficient. If they are able to do 75 J of useful work, what is their total work or energy input? Round your answer to the tens place. Be sure to include the appropriate unit in the answer. .67=75/x

OpenStudy (anonymous):

I tried to divide but I think i'm wrong

OpenStudy (anonymous):

do you how to solve them

OpenStudy (shane_b):

Do you know the formula for calculating efficiency? We should start with that...

OpenStudy (anonymous):

Yea but I don't think I have the right answers

OpenStudy (shane_b):

\[E_{ff}=\frac{output}{input}\]Plugging in the values you were given you have:\[0.67=\frac{output}{130J}\]Now you just need to solve for output.

OpenStudy (anonymous):

Is the output the answer when you divide 67 by 130

OpenStudy (shane_b):

You need to take the equation above and isolate the output to one side...that doesn't involve division :)

OpenStudy (mertsj):

If a person is 67% efficient in lifting an object, how much work was done when the person input 130 J of energy? This answer should be rounded to the ones place. It also needs to include the appropriate unit with the answer. .67=x/130 x=.67(130)=87.1

OpenStudy (shane_b):

Sorry @Mertsj, didn't know you were still helping

OpenStudy (anonymous):

Well I need all the help I can get so thanks to both of you

OpenStudy (mertsj):

A person is 67% efficient. If they are able to do 75 J of useful work, what is their total work or energy input? Round your answer to the tens place. Be sure to include the appropriate unit in the answer. .67=75/x .67x=75 x=75/.67 x= 111.9

OpenStudy (mertsj):

@Shane_B No problem. I didn't know that the asker did not know how to solve algebra 1 equations.

OpenStudy (shane_b):

Both answers involve simply using the efficiency formula and solving for a single unknown value...as @Mertsj is showing you

OpenStudy (anonymous):

I see them and yes I have a tough time with math

OpenStudy (anonymous):

Thanks @Shane_B and @Mertsj

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