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USPshnik [31]
3 years ago
11

Deduce a value for the latent heat of evaporation Lv of water. State clearly any simplifying assumptions that you make. Estimate

the pressure at which ice and water are in equilibrium at −2◦C given that ice cubes float with 4/5 of their volume submerged in water at the triple point (0.01◦C, 612Pa). [Latent heat of sublimation of ice at the triple point, Ls = 2776 °´ 103 Jkg−1.]

Physics
2 answers:
Stolb23 [73]3 years ago
5 0

Find the below attachment

azamat3 years ago
4 0

Answer:

Latent Heat of evaporation is L= 44.4 kJ/mol   and

The Pressure is 233.75 x 10^6 Pa

Explanation:

Explanation is in the following attachment.

                   

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What is acceleration
kap26 [50]

Answer:

The rate of change of velocity is acceleration. It's SI unit is m/s².

6 0
2 years ago
See the person on the right side of the front car. Six reference points could be used to show that the person is in / is NOT in
Rasek [7]

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1. According to person standing on ground ~

  • The person is in motion

2. According to The car ~

  • The person is not in motion

3. According to the Seat ~

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7 0
2 years ago
When comparing the three atomic models provided, which statement best describes the diagram?
aliina [53]
C. Three isotopes of the same element because the number of electrons and protons remain the same but the number of neutrons in the nucleus are different. This means that the isotopes have different masses to each other but the same chemical properties.
6 0
3 years ago
) Music. When a person sings, his or her vocal cords vibrate in a repetitive pattern that has the same frequency as the note tha
vaieri [72.5K]

(a) 0.0021 s, 2926.5 rad/s

The frequency of the B note is

f= 466 Hz

The time taken to make one complete cycle is equal to the period of the wave, which is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{466 Hz}=0.0021 s

The angular frequency instead is given by

\omega = 2\pi f

And substituting

f = 466 Hz

We find

\omega = 2\pi (466 Hz)=2926.5 rad/s

(b) 20 Hz, 125.6 rad/s

In this case, the period of the sound wave is

T = 50.0 ms = 0.050 s

So the frequency is equal to the reciprocal of the period:

f=\frac{1}{T}=\frac{1}{0.050 s}=20 Hz

While the angular frequency is given by:

\omega = 2\pi f = 2 \pi (20 Hz)=125.6 rad/s

(c) 4.30\cdot 10^{14} Hz, 7.48\cdot 1^{14} Hz, 2.33\cdot 10^{-15} s, 1.34\cdot 10^{-15}s

The minimum angular frequency of the light wave is

\omega_1 = 2.7\cdot 10^{15}rad/s

so the corresponding frequency is

f=\frac{\omega}{2 \pi}=\frac{2.7\cdot 10^{15}rad/s}{2\pi}=4.30\cdot 10^{14} Hz

and the period is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{4.30\cdot 10^{14}Hz}=2.33\cdot 10^{-15}s

The maximum angular frequency of the light wave is

\omega_2 = 4.7\cdot 10^{15}rad/s

so the corresponding frequency is

f=\frac{\omega}{2 \pi}=\frac{4.7\cdot 10^{15}rad/s}{2\pi}=7.48\cdot 10^{14} Hz

and the period is the reciprocal of the frequency:

T=\frac{1}{f}=\frac{1}{7.48\cdot 10^{14}Hz}=1.34\cdot 10^{-15}s

(d) 2.0\cdot 10^{-7}s, 3.14\cdot 10^{7} rad/s

In this case, the frequency is

f=5.0 MHz = 5.0 \cdot 10^6 Hz

So the period in this case is

T=\frac{1}{f}=\frac{1}{5.0\cdot 10^6  Hz}=2.0 \cdot 10^{-7} s

While the angular frequency is given by

\omega = 2\pi f=2 \pi (5.0\cdot 10^{6}Hz)=3.14\cdot 10^{7} rad/s

7 0
3 years ago
Find the rms speed of the molecules of a sample of n2 (diatomic nitrogen) gas at a temperature of 31.5°c.
Artemon [7]
The rms speed can be calculated using the following rule:
rms = sqrt ((3RT) / (M)) where:
R is the gas constant =  8.314 J/mol-K
T is the temperature = 31.5 + 273 = 304.5 degrees kelvin
M is the molar mass = 2*14 = 28 grams = 0.028 kg

Substitute with the givens to get the rms speed as follows:
rms speed = sqrt [(3*8.314*304.5) / (0.028)] = 520.811 m/sec
8 0
3 years ago
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