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Reil [10]
1 year ago
6

What fraction of the water must evaporate to remove precisely enough energy to keep the temperature constant? water at 37°c has

a latent heat of vaporization of lv = 580 kcal/kg.
Physics
1 answer:
mart [117]1 year ago
5 0

The fraction of the water must evaporate to remove precisely enough energy to keep the temperature constant when water at 37°c has a latent heat of vaporization of lv = 580 kcal/kg is 2.58 times 10 to the minus 3.

Vaporization is the process by which a substance is transformed from its liquid or solid state into its gaseous (vapour) state. Boiling is the term for the vaporization process when conditions permit the creation of vapour bubbles within a liquid. Sublimation is the process of directly converting a solid to a liquid.

Boiling and evaporation are the two processes that cause vaporization. Evaporation is the process by which a liquid body's surface changes from a liquid to a gas, as in the case of a drop of water on hot concrete evaporating into a gas. A liquid is said to be boiling when it is heated to the point at which it begins to give off steam, as when you boil water on a stove. The process of converting a substance from its liquid or solid state into its gaseous (vapour) state is known as vaporization.

To learn more about vaporization please visit - brainly.com/question/12625048
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Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du
ELEN [110]

With constant angular acceleration \alpha, the disk achieves an angular velocity \omega at time t according to

\omega=\alpha t

and angular displacement \theta according to

\theta=\dfrac12\alpha t^2

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}

b. Under constant acceleration, the average angular velocity is equivalent to

\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2

where \omega_f and \omega_i are the final and initial angular velocities, respectively. Then

\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}

c. After 1.00 s, the disk has instantaneous angular velocity

\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}

d. During the next 1.00 s, the disk will start moving with the angular velocity \omega_0 equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle \theta according to

\theta=\omega_0t+\dfrac12\alpha t^2

which would be equal to

\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}

5 0
3 years ago
an object is thrown with an initial horizontal velocity of 10 meters per second and take approximately 9 seconds to reach the gr
pantera1 [17]

The horizontal velocity was constant, so:

s = vt

s = 10\cdot9

s = 90

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Explanation:

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h = 0.52 m

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