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3 years ago

Which law of thermodynamics would be violated if heat were to spontaneously flow between two objects of equal temperature?

1 answer:

maks197457 [2]3 years ago

5 0

<span>Actually the second law of thermodynamics would truly gets violated ie, which means that the entrophy changes of the isolated system can never be negative, which covers the above that if heat were to spontaneously flow between any two objects of equal temperature would be fully violated.</span>

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The diagram shows the flow of water through a hydroelectric power station. The falling water

KonstantinChe [14]

Answer:

KE = $\frac{1}{2}$ m $v^{2}$

Explanation:

In the generation of energy from hydroelectric power station, the motion of water, and the turbines are paramount. The falling flowing water turns the blades of the turbine, which in-turn causes the movement of a coil within a strong magnetic field.

The motion of the coil which cuts the strong magnetic field induces current. Thus, the system generates electrical energy.

The equation that links kinetic energy (KE), mass (m) and speed (v) can be expressed as:

KE = $\frac{1}{2}$ m $v^{2}$

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2 years ago

Use the graph below to answer the following question: What is happening to the object's velocity?

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The object's velocity is decreasing.

Explanation:

From graph is the attached image, we can clearly point that the velocity of this motion is decreasing with time.

Velocity is a vector quantity.

The y-axis represent displacement.
The x-axis depicts time
Using the graph, we know that the slope of the line on the graph gives us the velocity as it denotes the change of displacement with time.
When we find the slope, it will give us a negative value which shows that the body is slowing down and not increasing speed.

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Velocity brainly.com/question/4460262

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3 years ago

Dale skis down a hill with a slope of 30°. Given that there is friction acting

Zarrin [17]

Answer:

The answer is A.

Explanation:

8 0

2 years ago

Read 2 more answers

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words: