Given:
F = ax
where
x = distance by which the rubber band is stretched
a = constant
The work done in stretching the rubber band from x = 0 to x = L is
![W=\int_{0}^{L} Fdx = \int_{0}^{L}ax \, dx = \frac{a}{2} [x^{2} ]_{0}^{L} = \frac{aL^{2}}{2}](https://tex.z-dn.net/?f=W%3D%5Cint_%7B0%7D%5E%7BL%7D%20Fdx%20%3D%20%5Cint_%7B0%7D%5E%7BL%7Dax%20%5C%2C%20dx%20%3D%20%5Cfrac%7Ba%7D%7B2%7D%20%20%5Bx%5E%7B2%7D%20%5D_%7B0%7D%5E%7BL%7D%20%3D%20%20%5Cfrac%7BaL%5E%7B2%7D%7D%7B2%7D%20)
Answer:
A red ladybug appears red in white light, red in red light, and black in blue light. Those would be the proper selections you'd need.
You may look at what group they are in
Group
1A=Group 1
2A = Group 2
3A = Group 13
4A= Group 14
5A=Group 15
6A=Group 16
7A=Group 17
The #A tells you how many valence electrons there are by the # before A. Such as Chlorine, which is in 7A, so therefore has 7 valence electrons.
-- The speed of light in air is very close to 3 x 10⁸ m/s.
Whatever the actual number is, it's equivalent to roughly
7 times around the Earth in 1 second. So for this kind of
problem, you can assume that we see things at the same time
that they happen; don't bother worrying about how long it takes
for the light to reach you.
-- For sound, it's a different story. Sound in air only travels at
about 340 m/s. It takes sound almost 5 seconds to go 1 mile.
-- Now, the lightning and thunder happen at the same time.
The light travels to you at the speed of light, so you see the
lightning pretty much when it happens. But the sound of the
thunder comes poking along at 340 m/s, and arrives AFTER
the sight of the lightning.
The length of time between the sight and the sound is about
99.9999% the result of the time it takes the sound to reach you.
If the thunder arrived at you 3 seconds after the light did, then
the sound traveled
(340 m/s) x (3 s) = 1,020 meters .
(about 0.63 of a mile)
(If you're worried about ignoring the time it takes
for the light to reach you ...
It takes light 0.0000034 second to cover the same 1,020 meters,
so including it in the calculation would not change the answer.)
Answer:
a) 35.44 mm
b) 17.67 mm
Explanation:
u = Object distance = 3.6 m
v = Image distance
f = Focal length = 35 mm
= Object height = 1.8 m
a) Lens Equation
The CCD sensor is 35.34 mm from the lens
b) Magnification
The person appears 17.67 mm tall on the sensor
