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shusha [124]
3 years ago
9

What energy transformations take place in Joule's system?

Physics
2 answers:
GREYUIT [131]3 years ago
5 0
The steam turns the turbine of an electrical generator, and the heat energy is transformed into the kinetic energy of the spinning turbine. The generator spins coils of wire in a magnetic field, transforming its kinetic energy into electrical energy.
IceJOKER [234]3 years ago
4 0

The gravitational potential energy of the weight changes to motion energy when the weight moves down and the paddle wheel spins. The motion energy changes to thermal energy, causing the water’s temperature to rise.

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Answer:

5. All of the statements are true; non is false

Explanation:

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The mass of an electron is _______. What is it?
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If a mass of a neutron is 1 the electron mass is 0.00054386734 and it's charge is negative. Hope this helps! ;)
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The Astronomer Carl Sagan is famous. Why?
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3 years ago
A compact disc (CD) stores music in a coded pattern of tiny pits 10−7m deep. The pits are arranged in a track that spirals outwa
andreev551 [17]

(a) 50 rad/s

The angular speed of the CD is related to the linear speed by:

\omega=\frac{v}{r}

where

\omega is the angular speed

v is the linear speed

r is the distance from the centre of the CD

When scanning the innermost part of the track, we have

v = 1.25 m/s

r = 25.0 mm = 0.025 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.025 m}=50 rad/s

(b) 21.6 rad/s

As in part a, the angular speed of the CD is given by

\omega=\frac{v}{r}

When scanning the outermost part of the track, we have

v = 1.25 m/s

r = 58.0 mm = 0.058 m

Therefore, the angular speed is

\omega=\frac{1.25 m/s}{0.058 m}=21.6 rad/s

(c) 5550 m

The maximum playing time of the CD is

t =74.0 min \cdot 60 s/min = 4,440 s

And we know that the linear speed of the track is

v = 1.25 m/s

If the track were stretched out in a straight line, then we would have a uniform motion, therefore the total length of the track would be:

d=vt=(1.25 m/s)(4,440 s)=5,550 m

(d) -6.4\cdot 10^{-3} rad/s^2

The angular acceleration of the CD is given by

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f = 21.6 rad/s is the final angular speed (when the CD is scanned at the outermost part)

\omega_i = 50.0 rad/s is the initial angular speed (when the CD is scanned at the innermost part)

t=4440 s is the time elapsed

Substituting into the equation, we find

\alpha=\frac{21.6 rad/s-50.0 rad/s}{4440 s}=-6.4\cdot 10^{-3} rad/s^2

5 0
3 years ago
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Becuz when you wash up in the tub you want layers of soap so you don’t stink
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