In what type of wave is the vibration perpendicular to the direction of travel of the wave
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Answer
Pressure, P = 1 atm
air density, ρ = 1.3 kg/m³
a) height of the atmosphere when the density is constant
Pressure at sea level = 1 atm = 101300 Pa
we know
P = ρ g h
h = 7951.33 m
height of the atmosphere will be equal to 7951.33 m
b) when air density decreased linearly to zero.
at x = 0 air density = 0
at x= h ρ_l = ρ_sl
assuming density is zero at x - distance
now, Pressure at depth x
integrating both side
now,
h = 15902.67 m
height of the atmosphere is equal to 15902.67 m.
A) See ray diagram in attachment (-6.0 cm)
By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:
where
q is the distance of the image from the lens
f = -10 cm is the focal length (negative for a diverging lens)
p = 15 cm is the distance of the object from the lens
Solving for q,
B) The image is upright
As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:
where are the size of the image and of the object, respectively.
Since q < 0 and p > o, we have that , which means that the image is upright.
C) The image is virtual
As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.
This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual
