Question

Please do this physics question im desperate!!! 25 points and brainliest!

Answer 1

Answer:

Given:

An electric motor uses 854 J of energy to lift 1.01 kg object to a height of 2.39 m.

To find

Efficiency of the motor

Concept:

The motor utilises energy to take the object to a certain height. Since gravitational force is an conservative force, this energy (in ideal conditions) gets stored in the object as Gravitational Potential Energy.

Calculation:

Gravitational Potential Energy of the object at that height

= m × g × h

= 1.01 × 10 × 2.39

= 24.139 J

But the energy utilised by the motor = 854 J.

So the rest amount of energy is lost as heat energy (via radiation losses)

So efficiency

={ (24.139)/854 } × 100%

= 2. 8265 %.

Answer 2

Question :-

What is the efficiency of an electron motor that uses 854 J of energy to lift a 1.01 kg object 2.39 m?

Answer :-

$\rightarrow \boxed{ \sf{ \pink{efficiency}\: } = \mathfrak{ \green{2.8265 \: \%}}} \:$

To Find :-

→ Efficiency of electron.

Formula used :-

$\sf{ \red{ gravitational \: potential \: energy }\: = mgh}$

Step - by - step explanation :-

Given that ,

• Height of object (h) = 2.39 m

• mass of object (m) = 1.01 kg

• used energy = 854 J

• assuming g = 10 m/s^{2}

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Solution :-

Substitute the given values in the given formula for gravitational potential energy,

$\implies \mathfrak{ \red{1.01 \times 10 \times 2.39} }\\ \\ \implies \mathfrak{ 24.139 \: j}$

Motor uses 854 J of energy ,

Therefore ,

$\rightarrow \small \boxed{ \sf{efficiency \: = \frac{potential \: energy}{used \: energy} \times 100} }\\ \\ \rightarrow \: \sf{ \red{efficiency }\: } = \mathfrak{ \green{\frac{24.139}{854} \times 100} }\\ \\ \rightarrow \sf{ efficiency \: } = \mathfrak{ \blue{ \frac{24139}{8540}} } \\ \\ \rightarrow \boxed{ \sf{ \green{efficiency}\: } = \mathfrak{ \red{ 2.8265 \: \%}}}$

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