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tamaranim1 [39]
3 years ago
5

An ice cube at 0°C is placed in a very large bathtub filled with water at 30°C and allowed to melt, causing no appreciable chang

e
in the temperature of the bath water. Which one of the following statements is true?
A. The net entropy change of the system (ice plus water) is zero because no heat was added to the system
B. The entropy lost by the ice cube is equal to the entropy gained by the water
C. The entropy of the system (ice plus water) increases because the process is irreversible
D. The entropy of the water does not change because its temperature did not change
E. The entropy gained by the ice cube is equal to the entropy lost by the water​
Physics
1 answer:
Paladinen [302]3 years ago
6 0

Answer:

C. The entropy of the system (ice plus water) increases because the process is irreversible.

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A small bug is lying resting when nine ants simultaneously ambush it and begin pulling it in different directions. They are each
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Answer: The bug will remain motionless

Explanation:

According to Newton's first Law of Motion (sometimes called Law of Inertia):

<em>An object at rest or describing a uniform straight line motion (moving at constant velocity), will remain at rest or moving unless an external force is applied to it and changes its state of rest or motion. </em>

In other words:

An object or body will keep its state of motion until an external force changes its state

This means that objects tend to remain in its state of motion, and is the definition of the inertia, as well.

In addition, according to his law, an object in rest can be in equilibrium (net force equals to zero), and a moving object can also be in equilibrium, as long as it keeps a constant velocity.

<h2>This is why the bug, which is at rest will remain at rest, although the ants are simultaneously pulling it in different directions, since the resultant of all these forces is zero.</h2>
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3 years ago
The power grid is the system that transfers electricity to homes and businesses. A power plant produces electricity. The output
sasho [114]

Answer: energy

Explanation: So the input and output of the power grid system is energy.

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In a series lrc circuit, the frequency at which the circuit is at resonance is f0. If you double the resistance, the inductance,
taurus [48]

When you double capacitance and inductance, the new resonance frequency becomes f/2.

  • Resonance frequency:

The resonance frequency of RLC series circuit, is the frequency at which the capacity reactance is equal to inductive reactance.

It can also be defined as the natural frequency of an object where it tends to vibrate at a higher amplitude.

Xc = Xl

which gives the value for resonance frequency:

f = \frac{1}{2\pi \sqrt{LC} }

where;

f is the resonance frequency

L is the inductance

C is the capacitance

When you double capacitance and inductance, the new resonance frequency becomes;

f' = \frac{1}{2\pi \sqrt{2L2C} }

f' = \frac{1}{2\pi \sqrt{4LC} }

f' = \frac{1}{\pi \sqrt{LC} }\frac{1}{2}

f' = \frac{1}{2} f

Thus from above,

When you double capacitance and inductance, the new resonance frequency becomes f/2.

Learn more about resonance frequency here:

<u>brainly.com/question/13040523</u>

#SPJ4

6 0
2 years ago
A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri
GREYUIT [131]

Answer:

|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}

Now, substitute this into 'dF':

d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)

Now we can integrate dF over the rod.

\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)

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