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mylen [45]
2 years ago
9

G 2.3 mol of monatomic gas a initially has 4800 j of thermal energy. it interacts with 2.9 mol of monatomic gas b, which initial

ly has 8500 j of thermal energy. part a which gas has the higher initial temperature?
Physics
1 answer:
77julia77 [94]2 years ago
4 0
To determine the gas which has the higher initial temperature, we need an equation that would relate energy and temperature. From thermodynamics, we use the expression Energy = nCvT where n is the number of moles, Cv is the heat capacity at constant volume and T is the temperature. By evaluating the temperature of both gases we determine the which would have higher temperature.

4800 = 2.3CvT
T = 2086.96/Cv

8500 = 2.9CvT
T = 2931.03/Cv

Assuming that both gases are the same. Therefore, the value of Cv for both would be equal. So, we conclude that it is the second gas with 2.9 moles would have the higher initial temperature.
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Charge is distributed uniformly on the surface of a large flat plate. the electric field 2 cm from the plate is 30 n/c. the elec
AysviL [449]
The electric field produced by a large flat plate with uniform charge density on its surface can be found by using Gauss law, and it is equal to
E= \frac{\sigma}{2\epsilon_0}
where
\sigma is the charge density
\epsilon_0 is the vacuum permittivity

We see that the intensity of the electric field does not depend on the distance from the plate. Therefore, the strenght of the electric field at 4 cm from the plate is equal to the strength of the electric field at 2 cm from the plate:
E=30 N/C
7 0
2 years ago
If there is a potential difference v between the metal and the detector, what is the minimum energy emin that an electron must h
beks73 [17]
The electrical potential energy of a charge q located at a point at potential V is given by
U=qV
Therefore, if the charge must move between two points at potential V1 and V2, the difference in potential energy of the charge will be
\Delta U = q (V_2 -V_1)=q \Delta V

In our problem, the electron (charge e) must travel across a potential difference V. So the energy it will lose traveling from the metal to the detector will be equal to 
\Delta U = e V
Therefore, if we want the electron to reach the detector, the minimum energy the electron must have is exactly equal to the energy it loses moving from the metal to the detector:
E_{min} = \Delta U = eV
5 0
2 years ago
Although blood cells are contained within a special liquid called plasma, the cells themselves are___________.
valentina_108 [34]

Answer:

Solid

Explanation:

The plasma is the liquid part of blood, it is 90% and accounts for 55% of blood volume. It is what red blood cells, white blood cells, and platelets move around in. These cells remain solid within the plasma. I hoped this helped!

8 0
2 years ago
A thick steel sheet of area 100 in.2 is exposed to air near the ocean. After a one-year period it was found to experience a weig
vladimir1956 [14]

Answer:

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

Explanation:

The corrosion rate is the rate of material remove.The formula for calculating CPR or corrosion penetration rate is

CPR=\frac{KW}{DAT}

K= constant depends on the system of units used.

W= weight =485 g

D= density =7.9 g/cm³

A = exposed specimen area =100 in² =6.452 cm²

K=534 to give CPR in mpy

K=87.6  to give CPR in mm/yr

mpy

CPR=\frac{KW}{DAT}

        =\frac{534\times( 485g)\times( 10^3mg/g)}{(7.9g/cm^3) \times (100in^2)\times (24h/day)\times (365day/yr)\times 1yr}

        =37.4mpy

mm/yr

CPR=\frac{KW}{DAT}

        =\frac{87.6\times (485g)\times (10^3 mg/g)}{(7.9g/cm^3)\times (100in^2)\times(2.54cm/in)^2\times (24h/day)\times (365day/yr)\times 1yr}

       =0.952 mm/yr

Therefore the rate of corrosion 37.4 mpy and 0.952 mm/yr.

3 0
2 years ago
Pendulum clocks are made to run at the correct rate by adjusting the pendulum’s length. Suppose you move from one city to anothe
levacccp [35]

Answer:

Obviously Lengthen...   T = 2\pi \sqrt{L/g}   or   g = 4\pi ^{2} L/g

Explanation:

As we can observe from the equation, time period of a simple pendulum depends upon the length directly. When the gravitational acceleration increases the time period of the pendulum decreases and vice versa. So, by increasing the length, the time period can be adjusted...

4 0
2 years ago
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