Answer:
A. The volume occupied by the molecules can cause an increase in pressure compared to the ideal gas.
D. As attractive forces between molecules increase, deviations from ideal behavior become more apparent at relatively low temperatures.
Explanation:
as we know by real gas equation
while ideal gas equation is given as
so from above formula we can say that net pressure is increased in real gas
Also we know that all real gas will close to behave like ideal gas when the pressure of the real gas is low and temperature of the gas is high
So above are the correct observations
<span>373.2 km
The formula for velocity at any point within an orbit is
v = sqrt(mu(2/r - 1/a))
where
v = velocity
mu = standard gravitational parameter (GM)
r = radius satellite currently at
a = semi-major axis
Since the orbit is assumed to be circular, the equation is simplified to
v = sqrt(mu/r)
The value of mu for earth is
3.986004419 Ă— 10^14 m^3/s^2
Now we need to figure out how many seconds one orbit of the space station takes. So
86400 / 15.65 = 5520.767 seconds
And the distance the space station travels is 2 pi r, and since velocity is distance divided by time, we get the following as the station's velocity
2 pi r / 5520.767
Finally, combining all that gets us the following equality
v = 2 pi r / 5520.767
v = sqrt(mu/r)
mu = 3.986004419 Ă— 10^14 m^3/s^2
2 pi r / 5520.767 s = sqrt(3.986004419 * 10^14 m^3/s^2 / r)
Square both sides
1.29527 * 10^-6 r^2 s^2 = 3.986004419 * 10^14 m^3/s^2 / r
Multiply both sides by r
1.29527 * 10^-6 r^3 s^2 = 3.986004419 * 10^14 m^3/s^2
Divide both sides by 1.29527 * 10^-6 s^2
r^3 = 3.0773498781296 * 10^20 m^3
Take the cube root of both sides
r = 6751375.945 m
Since we actually want how far from the surface of the earth the space station is, we now subtract the radius of the earth from the radius of the orbit. For this problem, I'll be using the equatorial radius. So
6751375.945 m - 6378137.0 m = 373238.945 m
Converting to kilometers and rounding to 4 significant figures gives
373.2 km</span>
Answer:
a)1.51*10^-22joules b) 1.89*10^-7m
Explanation:
Work done to stop the proton = the kinetic energy of the proton = 1/2 mv^2 = 1/2* 1.67*10^-27* 425*425 = 1.51* 10 ^ -22 joules
b) net force acting to stop the proton = 8.01*10^-16
Work done needed to stop the proton = net force acting opposite the motion * distance
Distance covered = need work done/ net force
Distance = 1.51*10^-22/8.01*10^-16= 1.89*10^-7m
In order to read the publications of his peers, or read his own notes of the work
that he did on the previous day, or find his coffee mug on his desk in the lab, the
research scientist must arrange to have each of them illuminated with visible
wavelengths of light, and then he must catch the light reflected from each of them
with his eyes.
