V=(40km/hr)(hr/3600s)(1000000mm/km)
v=11111.1mm/s
v=d/t
d=vt
d=(11111.1mm/s)(5s)
d=55555mm
d=5.56x10^4mm
To solve the problem, it is necessary the concepts related to the definition of area in a sphere, and the proportionality of the counts per second between the two distances.
The area with a certain radius and the number of counts per second is proportional to another with a greater or lesser radius, in other words,
M,m = Counts per second
Our radios are given by
Therefore replacing we have that,
Therefore the number of counts expect at a distance of 20 cm is 19.66cps
Answer:
the color is green
- 602.93 nm ( orange color )
the observation is that there is a change of visible color
Explanation:
A) wavelength of visible light that is most strongly reflected from a point on a soap
refraction n = 1.33
wall thickness (t) = 290 nm
2nt = (2m +1 ) ∝/2 -----equation 1
note when m = 0
therefore ∝ = 4nt/ 1 = 4 * 1.33 * 290 = 1542.8nm we will discard this
when m = 1
equation 1 becomes
∝ = 4nt/3 =( 4 * 1.33 * 290) / 3 = 1542.8 / 3 = 514.27 ( wavelength )
the color is green
B) the wavelength when the wall thickness is 340 nm
∝ = 4nt / 2m +1
where m = 1
∝ = (4 * 1.33 * 340 ) / 3 = 1808.8 / 3 = 602.93 nm ( orange color )
the observation is that there is a change of visible color
Humid air has higher pressure because of the heaviness of the water
Explanation:
An perfect mass less spring, attached at one end and with a free mass attached at the other end, will have a distinct frequency of oscillation depending on its constant spring and mass. On the other hand, a spring with mass along its length will not have a characteristic frequency of oscillation.
Alternatively, based on its spring constant and mass per length, it will now have a wave Speed. It would be possible to use all wavelengths and frequencies, as long as the component fλ= S, where S is the spring wave size. If that sounds like longitudinal waves, like solid sound waves.
