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3 years ago

Based on the second law of thermodynamics why must a machine always be less than 100% efficient?

Physics

1 answer:

AleksandrR [38]3 years ago

4 0

Answer: B.Heat is never converted completely into mechanical energy.

Explanation:

According to the second principle of thermodynamics:

"The amount of entropy in the universe tends to increase over time"

However the first formulation of this law (by Sadi Carnot) states:

"There is an upper limit to the efficiency of conversion of heat to work, in a heat engine "

This means the heat cannot be completely transformed into mechanical energy in a machine. That is why a machine's effieciency is always less than 100%

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Butter is an example for liquid in solid or solid in liquid

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Answer:

Liquid in solid.

Explanation:

Butter is colloidal in nature belonging to the category of gels. This means that the dispersed phase is liquid and the dispersing medium is solid. In short, butter is an example for liquid in solid.

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3 years ago

A simple pendulum is used to determine the acceleration due to gravity at the surface of a planet. The pendulum has a length of

SVEN [57.7K]

Answer:

Acceleration due to gravity is 20 $m/sec^2$

So option (E) will be correct answer

Explanation:

We have given length of the pendulum l = 2 m

Time period of the pendulum T = 2 sec

We have to find acceleration due to gravity g

We know that time period of pendulum is given by

$T=2\pi \sqrt{\frac{l}{g}}$

$2=2\times 3.14 \sqrt{\frac{2}{g}}$

$0.3184= \sqrt{\frac{2}{g}}$

Squaring both side

$0.1014= {\frac{2}{g}}$

$g=19.71=20m/sec^2$

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So option (E) will be correct answer.

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3 years ago

Two workers are sliding 350 kgkg crate across the floor. One worker pushes forward on the crate with a force of 390 NN while the

svetoff [14.1K]

Answer:

$\mu_k=0.18$

Explanation:

First, we write the equations of motion for each axis. Since the crate is sliding with constant speed, its acceleration is zero. Then, we have:

$x: T+F-f_k=0\\\\y:N-mg=0$

Where T is the tension in the rope, F is the force exerted by the first worker, f_k is the frictional force, N is the normal force and mg is the weight of the crate.

Since $f_k=\mu_k N$ and $N=mg$ , we can rewrite the first equation as:

$T+F-\mu_k mg=0$

Now, we solve for $\mu_k$ and calculate it:

$\mu_k=\frac{T+F}{mg}\\ \\\mu_k =\frac{220N+390N}{(350kg)(9.8m/s^{2})} =0.18$

This means that the crate's coefficient of kinetic friction on the floor is 0.18.

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4 years ago