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Allisa [31]
3 years ago
9

A vertical cylinder with a heavy piston contains air at 300 K. The initial pressure is 2.0 x 105 Pa and the initial volume is 0.

35 m3 . Take the molar mass of air as 28.9 g/mol and assume Cv= 5 2R. (A) Find the specific heat of air at constant volume in units of J kg·K . (5 pts) (B) Calculate the mass of air in the cylinder. (5 pts) (C) Suppose the piston is fixed. Find the energy input required to raise the temperature to 700 K. (5 pts). (D) Assume again the conditions of the initial state and assume the piston is free to move. Find the energy input to raise the temperature to 700 K. (10 pts)

Physics
1 answer:
aleksklad [387]3 years ago
8 0

Find answers and explanations in the attachments

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How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n
grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

dSin\theta = m\lambda

Where

m= Any integer which represent the number of repetition of spectrum

\lambda= Wavelength

d = Distance between the slits.

\theta= Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

d = \frac{m\lambda}{sin\theta }

d = \frac{2*536*10^{-9}}{sin(24)}

d = 2.63*10^{-6}m

Therefore the number of lines per millimeter would be given as

\frac{1}{d} = \frac{1}{2.63*10^{-6} }

\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})

\frac{1}{d} = 379.4 lines/mm

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

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25 Points PLEASE I need Help
olga55 [171]

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Explanation:

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laila [671]

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A fluid in an aquifer is 23.6 m above a reference datum, the fluid pressure (in gage pressure) is 4390 n/m2 and the flow velocit
Phoenix [80]

As per Bernuolli's Theorem total energy per unit mass is given as

\frac{P}{\rho} + \frac{1}{2}v^2 + gH = E

now from above equation

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v = 7.22 * 10^{-4} m/s

H = 23.6 m

now by above equation

\frac{4390}{0.999*10^3} + \frac{1}{2}*(7.22*10^{-4})^2 + 9.8*23.6 = E

E = 235.7 J/kg

Part B)

Now energy per unit weight

U = \frac{E}{g}

U = \frac{235.7}{9.8}

U = 24 m

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3 years ago
A clothes dryer uses about 9 amps of current from a 240 volt line. How much power does it use?
Basile [38]
 240 x 9 = power
power= 2160 watts


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Read 2 more answers
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