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zhuklara [117]
3 years ago
7

Imagine you transfer heat into a gas. This heat splits up in two different ways: Additional thermal energy making the gas hotter

and “fuel” for the gas to do work. Let’s suppose, 64% of the transferred heat transforms into thermal energy. How many percent of the initial heat can be used to do work?
Physics
1 answer:
navik [9.2K]3 years ago
7 0

Answer:

  • <u>36% of the initial heat can be used to do work.</u>

Explanation:

The initial heat is the 100%, i.e. it is all the heat that will be transferred.

It wil be transferred into two different kinds of energy:

  • additional thermal energy that will increase the thermal energy of the gas, making it hotter. Call it U. U = 64%. And,

  • "fuel" for the gas to do work. Call it W.

Then,

  • 100% = U + W
  • W = 100% - U
  • W = 100% - 64% = 36%

That is, 36% of the initial heat is available to do work.

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You stand on a frictional platform that is rotating at 1.8 rev/s. Your arms are outstretched, and you hold a heavy weight in eac
dusya [7]

Answer:

20.62361 rad/s

489.81804 J

Explanation:

I_i = Initial moment of inertia = 9.3 kgm²

I_f = Final moment of inertia = 5.1 kgm²

\omega_i = Initial angular speed = 1.8 rev/s

\omega_f = Final angular speed

As the angular momentum of the system is conserved

I_i\omega_i=I_f\omega_f\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{9.3\times 1.8}{5.1}\\\Rightarrow \omega_f=3.28235\ rev/s=3.28235\times 2\pi=20.62361\ rad/s

The resulting angular speed of the platform is 20.62361 rad/s

Change in kinetic energy is given by

\Delta K=\dfrac{1}{2}(I_f\omega_f^2-I_i\omega_i^2)\\\Rightarrow \Delta K=\dfrac{1}{2}(5.1\times (20.62361)^2-9.3\times (1.8\times 2\pi)^2)\\\Rightarrow \Delta K=489.81804\ J

The change in kinetic energy of the system is 489.81804 J

As the work was done to move the weight in there was an increase in kinetic energy

6 0
3 years ago
Determine the approximate force (N) used to pull a sled up a 400 m hill using 1900 J of work.
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The work done to pull the sled up to the hill is given by
W=Fd
where
F is the intensity of the force
d is the distance where the force is applied.

In our problem, the work done is W=1900 J and the distance through which the force is applied is d=400 m, so we can calculate the average force by re-arranging the previous equation and by using these data:
F= \frac{W}{d}= \frac{1900 J}{400 m} = 4.75 N \sim 5 N
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The answer completely depends on the number that belongs in the space before the word "microfarad".
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In creating an accurate scale model of our solar system, Lana placed Earth 1 foot from the Sun. The actual distance from Earth t
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Pluto would be placed 39.74 feet far from the Sun.

Astronomers use the gap between Earth and the sun, which is ninety-three million miles, as a new unit of measure called the Astronomical Unit.

Map scale refers to the connection (or ratio) between the space on a map and the corresponding distance on the ground. For example, on a 1:one hundred thousand scale map, 1cm at the map equals 1km on the ground.

The distance between the earth and solar, a = around 150 million km, is defined as one Astronomical Unit (AU). The radius of the solar, the solar is around 700,000 km.

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The bending of light when it changes media is called
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That's called "refraction".
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