they act as sunshades from the sun intensity, and filters
Well, there you have a very important principle wrapped up in that question.
There's actually no such thing as a real, actual amount of potential energy.
There's only potential <em><u>relative to some place</u></em>. It's the work you have to do
to lift the object from that reference place to wherever it is now. It's also
the kinetic energy the object would have if it fell down to the reference place
from where it is now.
Here's the formula for potential energy: PE = (mass) x (gravity) x (<em><u>height</u></em><u>)</u> .
So naturally, when you use that formula, you need to decide "height above what ?"
If you're reading a book while you're flying in a passenger jet, the book's PE is
(M x G x 0 meters) relative to your lap, (M x G x 1 meter) relative to the floor of the
plane, (M x G x 10,000 meters) relative to the ground, and maybe (M x G x 25,000 meters)
relative to the bottom of the ocean.
Let's say that gravity is 9.8 m/s² .
Then a 4kg block sitting on the floor has (39.2 x 0 meters) PE relative to the floor
it's sitting on, also (39.2 x 3 meters) relative to the floor that's one floor downstairs,
also (39.2 x 30 meters) relative to 10 floors downstairs, and if it's on the top floor of
the Amoco/Aon Center in Chicago, maybe (39.2 x 345 meters) relative to the floor
in the coffee shop that's off the lobby on the ground floor.
The boiling point of water, or any liquid, varies according to the surrounding atmospheric pressure. A liquid boils, or begins turning to vapor, when its internal vapor pressure equals the atmospheric pressure.
answer
no
Explanation:
I do not think that I would because even though its a conductor in the insulator I think it would insulate it before it will work (not sure if that makes sense)
Answer:
y = 2.74 m
Explanation:
The linear thermal expansion processes are described by the expression
ΔL = α L ΔT
Where α the thermal dilation constant for concrete is 12 10⁻⁶ºC⁻¹, ΔL is the length variation and ΔT the temperature variation in this case 20ªc
If the bridge is 250 m long and is covered by two sections each of them must be L = 125 m, let's calculate the variation in length
ΔL = 12 10⁻⁶ 125 20
ΔL = 3.0 10⁻² m
Let's use trigonometry to find the height
The hypotenuse Lf = 125 + 0.03 = 125.03 m
Adjacent leg L₀ = 125 m
cos θ = L₀ / Lf
θ = cos⁻¹ (L₀ / Lf)
θ = cos⁻¹ (125 / 125.03)
θ = 1,255º
We calculate the height
tan 1,255 = y / x
y = x tan 1,255
y = 125 tan 1,255
y = 2.74 m
