Vi=12m/s Vf=16m/s t=8s a=? a=Vf-Vi/t=16-12/8=4/8=1/2 a=0.5m/s^2
At the top of the mountain, when he tightens the cap onto the bottole, there is some water and some air inside the bottle. Then he brings the bottle down to the base of the mountain.
The pressure on the outside of the bottle is greater than it was when he put the cap on. If anything could get out of the bottlde, it would. But it can't . . . the cap is on too tight. So all the water and all the air has to stay inside, and anything that can get squished into a smaller space has to get squished into a smaller space.
The water is pretty much unsquishable.
Biut the air in there can be <em>COMPRESSED</em>. The air gets squished into a smaller space, and the bottle wrinkles in slightly.
help me w mine and ill try to help with yours
This is a Fraunhofer single slit experiment, where the light passing through the slit produces an interference pattern on the screen, and where the dark bands (minima of diffraction) are located at a distance of

from the center of the pattern. In the formula, m is the order of the minimum,

the wavelenght,

the distance of the screen from the slit and

the width of the slit.
In our problem, the distance of the first-order band (m=1) is

. The distance of the screen is D=86 cm while the wavelength is

. Using these data and re-arranging the formula, we can find a, the width of the slit:
Answer:
- I have fond the answer
Explanation:
but my camera doesn't work