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Marizza181 [45]
3 years ago
8

If the period of a given wave is 6 seconds what si the frequency of the wave

Physics
1 answer:
Veronika [31]3 years ago
3 0
Frequency =1/period
Freq= 1/6= 0.17 Hertz
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The answer is c cause it is not moving like the other 3 
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What are alkaline earth metals used for?
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C firework

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A sample of 20.0 moles of a monatomic ideal gas (γ = 1.67) undergoes an adiabatic process. The initial pressure is and the initi
Alexeev081 [22]
There are some missing data in the text of the exercise. Here the complete text:
"<span>A sample of 20.0 moles of a monatomic ideal gas (γ = 1.67) undergoes an adiabatic process. The initial pressure is 400kPa and the initial temperature is 450K. The final temperature of the gas is 320K. What is the final volume of the gas? Let the ideal-gas constant R = 8.314 J/(mol • K). "

Solution:

First, we can find the initial volume of the gas, by using the ideal gas law:
</span>pV=nRT
<span>where 
p is the pressure
V the volume
n the number of moles
R the gas constant
T the absolute temperature

Using the initial data of the gas, we can find its initial volume:
</span>V_i =  \frac{nRT_i}{p_i} = \frac{(20.0 mol)(8.31 J/molK)(450 K)}{4 \cdot 10^5 Pa} =0.187 m^3
<span>
Then the gas undergoes an adiabatic process. For an adiabatic transformation, the following relationship between volume and temperature can be used:
</span>TV^{\gamma-1} = cost.
<span>where </span>\gamma=1.67 for a monoatomic gas as in this exercise. The previous relationship can be also written as
T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}
where i labels the initial conditions and f the final conditions. Re-arranging the equation and using the data of the problem, we can find the final volume of the gas:
V_f = V_i  \sqrt[\gamma-1]{ \frac{T_i}{T_f} }=(0.187 m^3) \sqrt[0.67]{ \frac{450 K}{320 K} }=0.310 m^3 = 310 L
So, the final volume of the gas is 310 L.
5 0
3 years ago
You decide to visit Santa Claus at the north pole to put in a good word about your splendid behavior throughout the year. While
ANTONII [103]

To solve this problem it is necessary to apply the concepts related to the Rotational Force described from the equilibrium and Newton's second law.

When there is equilibrium, the Force generated by the tension is equivalent to the Force of the Weight. However in rotation, the Weight must be equivalent to the Centrifugal Force and the tension, in other words:

W = F_T + m\omega^2r_E

Where

\omega = \frac{2\pi}{T} \rightarrow Angular velocity is equal to the Period, at this case Earth's period

r_E = 6.371*10^6m \rightarrow Radius of the Earth

m = mass

F_T= Force of Tension

W = mg \rightarrow Newton's second law

Replacing and re-arrange to find the Tension we have,

F_T = W- \frac{W}{g} (\frac{2\pi}{T})^2r_E

F_T = W(1-(\frac{2\pi}{T})^2\frac{r_E}{g})

F_T = (505)(1-(\frac{2\pi}{24hours})^2\frac{6.371*10^6}{9.8})

F_T = (505)(1-(\frac{2\pi}{24hours(\frac{3600s}{1hour})})^2\frac{6.371*10^6}{9.8})

F_T = (505)(1-(\frac{2\pi}{86400})^2\frac{6.371*10^6}{9.8})

F_T = 503.26N

Therefore when Sneezy is on the equator he is in a circular orbit with a Force of tension of 503.26N

7 0
3 years ago
Two point charges are 3.00 cm apart. They are moved to a new separation of 2.00 cm. By what factor does the resulting mutual for
trasher [3.6K]

Answer:

By a factor of 9/4

Explanation:

Applying Coulomb's law,

F = kqq'/r²................... Equation 1

Assuming q and q' are the two point charges respectively.

Where k = coulomb's constant, r = distance between the charges.

When the point charges are 3.0 m apart,

F = kqq'/3²

F = kqq'/9.................... Equation 1

When they are moved to a new distance, 2.00 m

F' = kqq'/2²

F' = kqq'/4................. Equation 2.

Comparing equation 1 and equation 2.

F' = 9F/4

Hence the resulting mutual force change by a factor of 9/4

3 0
4 years ago
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