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Rudiy27
3 years ago
14

A scientist has two samples of a substance. Both samples have the same temperature. One sample has a mass of 10 g. The other sam

ple has a mass of 20 g. Compare the average kinetic energy and total kinetic energy of the particles in each sample.
Chemistry
1 answer:
Serga [27]3 years ago
5 0

A scientist has two samples of a substance. Both samples have the same temperature. The average kinetic energy of the two samples is the same but the total kinetic energy of the particle with a mass of 20 g is greater than the one with a mass of 10 g.

From the given information:

  • sample A = 10 g
  • sample B = 20 g

           The average kinetic energy is relative to the temperature and the temperature can be used as a determining factor when measuring the average kinetic energy. From the given question, it was stated that both samples have the same temperature, as such, we can infer that they have the same average kinetic energy.

           The total kinetic energy of an object is equal to the sum of the kinetic energies produced from every kind of motion. The total kinetic energy of a particle is affected by the speed of the motion and the mass of the sample. As such, the mass of the sample with 20 g will have a greater total kinetic energy than a mass with 10 g.

Therefore, from the above explanation, we can conclude that the mass of the average kinetic energy of the two samples is the same but the total kinetic energy of the particle with a mass of 20 g is greater than the one with a mass of 10 g.

Learn more about kinetic energy here:

brainly.com/question/12669551?referrer=searchResults

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Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

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