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Marta_Voda [28]
3 years ago
12

A gas is collected over water at 20C in a eudiometer tube until the pressure is equilibrated with the atmospherica pressure. Wha

t is the preussre of the dry gas if the barometric pressure is 771?
Physics
1 answer:
solong [7]3 years ago
6 0

Let's assume that the pressure given is 771 torr.

Answer:

747.44 torr

Explanation:

The eudiometer tube is a setup that collects gas in a liquid (generally water). After the thermal equilibrium, the gas becomes in equilibrium with the vapor of water above the liquid.

By Dalton's law, the total pressure of a mixture is the sum of the partial pressure of its components. The partial pressure is the pressure that the gas would have at the same conditions if it was alone. So, the pressure of the dry gas is its partial pressure.

The total pressure is the barometric pressure, and the partial pressure of the water vapor at 20°C is 23.56 torr (the value can be found at a thermodynamic table). So, the pressure of the dry gas (P) is:

771 = 23.56 + P

P = 771 - 23.56

P = 747.44 torr

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As we know that the time period of simple pendulum is inversely proportional to the square root of acceleration due to gravity, thus the time period becomes infinity.

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A train accelerates at -1.5m/s for 10 seconds if the train had an initial speed of 32 m/s what is it's new speed
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A. 17 m/s. let me know if it’s correct
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3 years ago
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Zarrin [17]

Answeill give it a go i guess

Explanation:

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         1758                          78.5                               258                2.4

         1758                          78.5                               258                2.4

         1758                          78.5                               258                2.4

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3 0
3 years ago
A wheel is rotating about an axis that is in the z direction The angular velocity ωz is 6.00 rad s at t 0 increases linearly wit
Amanda [17]

A) +1.67 rad/s^2

The angular acceleration of the wheel is given by

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

where

\omega_i = -6.00 rad/s is the initial angular velocity of the wheel (initially clockwise, so with a negative sign)

\omega_f = 4.00 rad/s is the final angular velocity (anticlockwise, so with a positive sign)

\Delta t= 6.00 s - 0=6.00 s is the time interval

Substituting into the equation, we find the angular acceleration:

\alpha = \frac{4.00 rad/s - (-6.00 rad/s)}{6.00 s}=+1.67 rad/s^2

And the acceleration is positive since the angular velocity increases steadily from a negative value to a positive value.

B) 3.6 s

The time interval during which the angular velocity is increasing is the time interval between the instant t_1 where the angular velocity becomes positive (so, \omega_i=0) and the time corresponding to the final instant t_2 = 6.0 s, where \omega_f = +6.00 rad/s. We can find this time interval by using

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

And solving for \Delta t we find

\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{+6.00 rad/s-0}{+1.67 rad/s^2}=3.6 s

C) 2.4 s

The time interval during which the angular velocity is idecreasing is the time interval between the initial instant t_1=0 when \omega_i=-4.00 rad/s) and the time corresponding to the instant in which the velovity becomes positive t_2, when \omega_f = 0 rad/s. We can find this time interval by using

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

And solving for \Delta t we find

\Delta t = \frac{\omega_f - \omega_i}{\alpha}=\frac{0-(-4.00 rad/s)}{+1.67 rad/s^2}=2.4 s

D) 5.6 rad

The angular displacement of the wheel is given by the equation

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where we have

\omega_i = -6.00 rad/s is the initial angular velocity of the wheel

\omega_f = 4.00 rad/s is the final angular velocity

\alpha=+1.67 rad/s^2 is the angular acceleration

Solving for \theta,

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{((+6.00 rad/s)^2-(-4.00 rad/s)^2}{2(+1.67 rad/s^2)}=5.6 rad

3 0
3 years ago
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