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.
Answer:
15.76N
Explanation:
horizontal component =Fcos¢ = 20cos38 = 15.76N
Answer:
v = 10 m/s
Explanation:
Given that,
Distance covered by a sprinter, d = 100 m
Time taken by him to reach the finish line, t = 10 s
We need to find his average velocity. We know that velocity is equal to the distance covered divided by time taken. So,
v = d/t
Hence, his average velocity is 10 m/s.
Answer:
Explanation:
Threshold frequency = 4.17 x 10¹⁴ Hz .
minimum energy required = hν where h is plank's constant and ν is frequency .
E = 6.6 x 10⁻³⁴ x 4.17 x 10¹⁴
= 27.52 x 10⁻²⁰ J .
wavelength of radiation falling = 245 x 10⁻⁹ m
Energy of this radiation = hc / λ
c is velocity of light and λ is wavelength of radiation .
= 6.6 x 10⁻³⁴ x 3 x 10⁸ / 245 x 10⁻⁹
= .08081 x 10⁻¹⁷ J
= 80.81 x 10⁻²⁰ J
kinetic energy of electrons ejected = energy of falling radiation - threshold energy
= 80.81 x 10⁻²⁰ - 27.52 x 10⁻²⁰
= 53.29 x 10⁻²⁰ J .
Answer:
weightlessness, condition experienced while in free-fall, in which the effect of gravity is canceled by the inertial (e.g., centrifugal) force resulting from orbital flight. ... Excluding spaceflight, true weightlessness can be experienced only briefly, as in an airplane following a ballistic (i.e., parabolic) path.
