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den301095 [7]
3 years ago
9

If you put a cup of hot chocolate at 90c on a table in a room kept at 25c you know it will cool down. How cool will the hot choc

olate get?
Physics
1 answer:
bagirrra123 [75]3 years ago
7 0

Answer:

Explanation:

Assuming we have a two hot liquid like 70°C tea and 30°C water, the tea will cool down but not to 30°C but to an equilibrium temperature say 40°C

But in this case, the temperature of the chocolate will drop almost to the room temperature eventually,

The heat in the chocolate will even out into the room temperature and the room will get slightly warmer until the are both in equilibrium temperature. But you won't notice this little change but if you go out of the room, you might notice that change...

So this little fraction is not always notice, so we will still considered it as the room temperature

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Answer:

The field gets weaker

Explanation:

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What changes when the force of gravity changes?
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The strength of the gravitational force between two objects depends on two factors, mass and distance. the force of gravity the masses exert on each other. If one of the masses is doubled, the force of gravity between the objects is doubled. increases, the force of gravity decreases.

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How would a coroner take a body's temperature?
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Answer: Rectal Thermometer

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4 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

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