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ahrayia [7]
3 years ago
13

How much heat is released when 432 g of water cools down from 71'c to 18'c?

Physics
1 answer:
maria [59]3 years ago
7 0
The heat released by the water when it cools down by a temperature difference \Delta T is
Q=mC_s \Delta T
where
m=432 g is the mass of the water
C_s = 4.18 J/g^{\circ}C is the specific heat capacity of water
\Delta T =71^{\circ}C-18^{\circ}C=53^{\circ} is the decrease of temperature of the water

Plugging the numbers into the equation, we find
Q=(432 g)(4.18 J/g^{\circ}C)(53^{\circ}C)=9.57 \cdot 10^4 J
and this is the amount of heat released by the water.
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A solid sphere of mass 4.0 kg and radius 0.12 m starts from rest at the top of a ramp inclined 15°, and rolls to the bottom. The
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v = 4.1 m/s

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so we will say

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now for pure rolling condition we know that

v = R\omega

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7 0
3 years ago
A wheel decelerates from 13.5 rad s−1 to 6.0 rad s−1 in 7 s. Calculate the angular displacement​
Igoryamba

Answer:

<em>Angular displacement=68.25 rad</em>

Explanation:

<u>Circular Motion</u>

If the angular speed varies from ωo to ωf in a time t, then the angular acceleration is given by:

\displaystyle \alpha=\frac{\omega_f-\omega_o}{t}

The angular displacement is given by:

\displaystyle \theta=\omega_o.t+\frac{\alpha.t^2}{2}

The wheel decelerates from ωo=13.5 rad/s to ωf=6 rad/s in t=7 s, thus:

\displaystyle \alpha=\frac{6-13.5}{7}

\displaystyle \alpha=\frac{-7.5}{7}

\displaystyle \alpha=-1.071 \ rad/s^2

Thus, the angular displacement is:

\displaystyle \theta=13.5*7+\frac{-1.071*7^2}{2}

\displaystyle \theta=94.5-26.25

\boxed{\displaystyle \theta=68.25\ rad}

Angular displacement=68.25 rad

3 0
3 years ago
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