Explanation:
let's assume that:
v1= 600ml=0,6l
T1=27°C= 300K
p1=700mmHG=93326Pa
T2=-20°C=253K
p2=500mmHg=66661
V2=?
p1V1/T1=p2V2/T2 => V2=p1V1T2/p2
V2= 93326*0,6*253/66661
V2=212,52l
Distance from the sun.
<span>The third law of planetary motion states that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit</span>. The semi-major axis is the distance from the sun to the epicenter of the ellipse (which would be the planet in question). So, the revolutionary period is directly related to the distance of the planet from the sun.
Answer:
I know one answer is an observer on earth sees a lunar eclipse
<span>2 Nitrogen, 4 Hydrogen, 3 Oxygen
9 atoms per molecule.
NH4, ammonium (not to be confused with ammonia NH3) is a 1+ ion and NO3 is 1-.
</span>
Answer:
A) 1.122 m
B) 5.4 hertz
C) 6.06 m/s ( towards negative x axis )
D) 0.22
E) max speed = amplitude * 34
= 0.22 * 34 = 7.48
minimum speed ( speed at rest ) = 0
Explanation:
The travelling wave = D = 0.22sin ( 5.6x + 34t)
this wave is represented in the form : D = Asin ( kx + wt)
A) wavelength
k = 2
/ v
v = wavelength
k = 5.6
therefore wavelength ( v ) from the equation = 2
/ 5.6 = 1.122 m
B ) frequency
w = 2
f
w = 34
f = frequency
therefore f = w / 2
= 34 / 2
= 5.4 hertz
C ) velocity
speed = f v
v = 5.4 * 1.122 = 6.06 m/s ( towards negative x axis )
D) amplitude
A = 0.22
E ) maximum and minimum speeds of particles on the chord
max speed = amplitude * 34
= 0.22 * 34 = 7.48
minimum speed ( speed at rest ) = 0