To solve this problem we will apply the definition of the ideal gas equation, where we will clear the density variable. In turn, the specific volume is the inverse of the density, so once the first term has been completed, we will simply proceed to divide it by 1. According to the definition of 1 atmosphere, this is equivalent in the English system to

The ideal gas equation said us that,
PV = nRT
Here,
P = pressure
V = Volume
R = Gas ideal constant
T = Temperature
n = Amount of substance (at this case the mass)
Then

The amount of substance per volume is the density, then

Replacing with our values,


Finally the specific volume would be


D is your answer hope this helps
Answer:
F = 4000 N
Explanation:
given,
mass of rocket (M)= 5000 Kg
10 Kg gas burns at speed (m)= 4000 m/s
time = 10 s
average force = ?
at the end the rocket is at rest
by conservation of momentum
M v + m v' = 0
5000 x v - 10 x 4000 = 0
5000 v = 40000
v = 8 m/s
speed of rocket = 8 m/s
now,
we know
change in momentum = F x Δ t


F = 4000 N
Hence, the average force applied to the rocket is equal to F = 4000 N
The hot gases produce their own characteristic pattern of spectral lines, which remain fixed as the temperature increases moderately.
<h3><u>Explanation: </u></h3>
A continuous light spectrum emitted by excited atoms of a hot gas with dark spaces in between due to scattered light of specific wavelengths is termed as an atomic spectrum. A hot gas has excited electrons and produces an emission spectrum; the scattered light forming dark bands are called spectral lines.
Fraunhofer closely observed sunlight by expanding the spectrum and a huge number of dark spectral lines were seen. "Robert Bunsen and Gustav Kirchhoff" discovered that when certain chemicals were burnt using a Bunsen burner, atomic spectra with spectral lines were seen. Atomic spectral pattern is thus a unique characteristic of any gas and can be used to independently identify presence of elements.
The spectrum change does not depend greatly on increasing temperatures and hence no significant change is observed in the emitted spectrum with moderate increase in temperature.
Answer:
a) 
b) 
Explanation:
Given:
height of water in one arm of the u-tube, 
a)
Gauge pressure at the water-mercury interface,:

we've the density of the water 


b)
Now the same pressure is balanced by the mercury column in the other arm of the tube:



<u>Now the difference in the column is :</u>


