**Physics Problems** play a special role in physics and it makes the physics just like an exciting game. But some students don't realize that. So, while dealing with physics, many of them have a feeling why do problems come? Understanding the physics theory and formula is enough why do we need problems? They even feel panic when problems part come. So, there is an **Excellent Solution** for this problem where you can get the solved problems in an systematic way in the form of steps that will make you analyze the problems.

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It is a great online tool which any student can use to learn physics.

Besides problem solvers, a lot of online sites and resources are also available to make physics simple.

With step-by-step explanations and solved and unsolved practice problems, they provide students with plenty of opportunities to hone their physics problem solving skills.

Practice and thorough understanding of concepts are the key to getting great grades in physics. This is where the physics equation solver makes things easy.

It is a great online tool which any student can use to learn physics.

We are solving the problems with the following steps which will help the student analyze the problem:

- Description
- Explanation
- Formula Display
- Diagram ID

(i) Calculate the efficiency of both engines A and B

(ii) Which Engine is more efficient?

(i) For Engine A,
T_{1} = 1000K,

T_{2} = 500K

$\eta$ = {1- $\frac{T_{2}}{T_{1}}$}

= {1- $\frac{500}{1000}$}

= 1 - 0.5

= 0.5

$\therefore$ The efficiency of engine A is 50%

(ii) For engine B, T_{1} = 500K

T_{2} = 400K

$\eta$ = {1- $\frac{T_{2}}{T_{1}}$}

= {1- $\frac{400}{500}$}

= 1 - 0.8

= 0.2

The efficiency of engine B is 20%

$\therefore$ engine**A** is more efficient compared to engine **B**.

T

$\eta$ = {1- $\frac{T_{2}}{T_{1}}$}

= {1- $\frac{500}{1000}$}

= 1 - 0.5

= 0.5

$\therefore$ The efficiency of engine A is 50%

(ii) For engine B, T

T

$\eta$ = {1- $\frac{T_{2}}{T_{1}}$}

= {1- $\frac{400}{500}$}

= 1 - 0.8

= 0.2

The efficiency of engine B is 20%

$\therefore$ engine

Given I = 10 A,

R = 20 $\Omega$

According to Ohms law, V = IR

where V = Voltage,

I = Current,

R = Resistance.

$\therefore$ V = 10 $\times$ 20

= 200V.

R = 20 $\Omega$

According to Ohms law, V = IR

where V = Voltage,

I = Current,

R = Resistance.

$\therefore$ V = 10 $\times$ 20

= 200V.