Simple physics question here. Get easy points.?

Question

Still helping my grandaughter with her studies. Can you answer these questions? Thankyou so much.


1.)A worker applies a force of 550N in sliding a block of wood 15.0m along a surface. The work output is determined to be 5.00x10^3 J. Calculate the workers efficiency.

2.)On a camping trip, the following scenario occurs:
I A camper rubs a match on a gritty surface igniting a match.
II The match is then used to ignited a pile of wood kindling, creating a bonfire
III While standing by the bonfire, the camper senses that the part of his body facing the fire is warm while his backside remains cool.
IV Another group of campers across the lake spots the bonfire.

Which of the above describes a conversion of mechanical energy to heat?

3.)Which of the above describes a conversion of radiant evergy heat? I, II, III OR IV?

Thankyou so much!

Thankyou I am so grateful.

Answer 1

I don't remember how to do 1 but her book will have the form you need. just remember to break J and N into it's units. ( i think N's are kgm/s^2 don't remember what J's are)
2 is l
3 is lll
Hope that helps

Answer 2

1. Work is Force times Distance

You have the *actual* physical work being done by sliding the block according to the formula. Then you have the bodily work done by the worker which is more. The efficiency is the ratio of the two amounts (output work / input work) (how much work was actually done, vs how much work was put into it).

Actual work = 550 N * 15 m = 2750 Joules
Worker work (muscles) = 5000 Joules

Efficiency = 2750 / 5000 = 0.55
(note: efficiency is *always* less than 1.00, otherwise you are breaking physical laws of thermodynamics.)

2. I . It's obvious to me -- the only mechanical energy being applied is in moving a match.

3. III . It could be II, but that is representative of convection, not radiation. IV is radiation also, but of the light, only -- not the heat.

.

Answer 3

For the first question,

efficiency = work output/ work input

work input = Force x distance

work input = 550 N x 15m = 8250 J

5000 J / 8250 J = 61%

For the second one,

Only I describes the conversion of mechanical energy to heat.

For the third one,
Situation III describes a conversion from radiant energy to heat.

Answer 4

I'm sorry that I can't remember the formula to answer the first qeustion. As for the second and third I believe that the answers are roman numerals I and II. If you can find the formula for efficiency you can just plug in the numbers and then do the math and you'll have the answer.

Answer 5

something is erroneous you ought to comprehend before everything that artwork as defined via a Physicist isn't comparable to the widespread meaning of artwork. artwork = stress situations displacement is the formulation Physicists prepare. So the artwork carried out via using 10 newtons of horizontal stress to shift an merchandise 4 metres for the period of a room demands 40 newton metres or 40 joules of artwork, if the stress is continuous and displacement is in an identical course through fact the stress utilized as on your occasion. whilst the utilized stress and the displacement vector are in distinctive instructions then the formulation to be utilized is Fd cos theta the place theta is the perspective between the stress vector and the displacement vector. on your situation theta would not be utilized on your equation as stress and displacement are in an identical course yet whilst it became it may equivalent a million in fee not 0.

Answer 6

i dont understand head to toe of the question
anyways thanks for the points!!!