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Brut [27]
3 years ago
13

A student does 25 J of work on the handle of a pencil sharpener. If the pencil sharpener does 20 J of work on the pencil, what i

s the efficiency of the sharpener?
Physics
1 answer:
allsm [11]3 years ago
3 0

Answer:

80%

Explanation:

Efficiency of machine = work output/work input  ×100 %

From question, work output = 20J

Work input = 25J.

Therefore efficiency = 20/25 × 100 %

Efficiency = 20×4 %

Efficiency = 80%

I hope this was helpful, please mark as brainliest

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Jefferson's administration was the beginning of popular government in the United States.
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He repealed federal taxes, Jefferson reduced military spending. He bought Louisiana from France. Considering only all these positive points and leaving aside the negative ones, one would say that the statement that "Jefferson's administration was the beginning of popular government in the United States." Is True!
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3 years ago
Acceleration and Force
olga55 [171]

Answer:

I'm pretty sure its 3m/s^2 for the acceleration but I don't know the force part sorry .

Explanation:

15m/s - 0m/s divided by 5 s = 3m/s

I'm no expert or anything so I could be wrong but this is the best I can give you. Sorry

6 0
2 years ago
A 3.0 kg block is pushed by a 14 N force. If µ = 0.6, will the block move?
Anna71 [15]

Answer:

The block will not move.

Explanation:

We'll begin by calculating the frictional force. This can be obtained as follow:

Coefficient of friction (µ) = 0.6

Mass of block (m) = 3 Kg

Acceleration due to gravity (g) = 10 m/s²

Normal reaction (R) = mg = 3 × 10 = 30 N

Frictional force (Fբ) =?

Fբ = µR

Fբ = 0.6 × 30

Fբ = 18 N

From the calculations made above, the frictional force of the block is 18 N. Since the frictional force (i.e 18 N) is bigger than the force applied (i.e 14 N), the block will not move.

4 0
3 years ago
An ideal refrigerator does 130. 0 j of work to remove 780. 0 j of heat from its cold compartment during each cycle. what is the
zheka24 [161]

The refrigerator's coefficient of performance is 6.

The heat extracted from the cold reservoir Q cold (i.e., inside a refrigerator) divided by the work W required to remove the heat is known as the coefficient of performance, or COP, of a refrigerator (i.e., the work done by the compressor). The required inside temperature and the outside temperature have a significant impact on the COP.

As the inside temperature of the refrigerator decreases, its coefficient of performance decreases. The coefficient of performance (COP) of refrigeration is always more than 1.

The heat produced in the cold compartment, H = 780.0 J

Work done in ideal refrigerator, W = 130.0 J

Refrigerator's coefficient of performance = H/W

                                                                     = 780/130

                                                                     = 6

Therefore, the refrigerator's coefficient of performance is 6.

Energy conservation requires the exhaust heat to be = 780 + 130

                                                                                          = 910 J

Learn more about  coefficient here:

brainly.com/question/18915846

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2 years ago
Too much skepticism can
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