I don't like any of those choices. But if you absolutely have to pick your answer
from this list, then it has to be 'D'.
The ocean is an enormous storage vessel for heat. It gets heat from the air in
the Summer ... which somewhat cools places near the coast ... and it releases
heat into the air during the Winter ... which warms places near the coast.
So I guess it's true that ocean surfaces change temperature more slowly than
land surfaces do, and they influence the land nearby in the process. But this
ignores the reason for the slow changes in ocean surface temperature. It's a
lot like saying that the loud noise produced by a race car is the result of the
car's ability to appear in a far different location after a short time.
In order to answer this exercise you need to use the formulas
S = Vo*t + (1/2)*a*t^2
Vf = Vo + at
The data will be given as
Vf = final velocity = ?
Vo = initial velocity = 1.4 m/s
a = acceleration = 0.20 m/s^2
s = displacement = 100m
And now you do the following:
100 = 1.4t + (1/2)*0.2*t^2
t = 25.388s
and
Vf = 1.4 + 0.2(25.388)
Vf = 6.5 m/s
So the answer you are looking for is 6.5 m/s
Answer:
it will take for the sphere to increase in potential by 1500 V, 503.71 s.
Explanation:
The charge on the sphere after t seconds is:
q = (1.0000049 - 1.0000000) t = 0.0000049 t
The voltage on the surface is
V = k * 
= k 0.0000049 t / R
solve for t
t = (R*V) / (0.0000049 k) = (0.12 * 1500) / (0.0000049 * 
) = 503.71 s
Answer:
The answer is 1250 mm
Explanation:
The path that particles move in the spectrometer is semicircular. Each of the particles has a displacement of twice the radius (2r) from the entrance to where the film is hit. According to the exercise, if the separation between the two molecules is 0.25 mm, then the difference in the radius of the molecules is equal to 0.125 mm. The ratio of mass/radius is equal for molecules, and therefore is equal to:
m = q * B * r / v
m/r = constant
(m/r)CO = (m/r)N = (28.0106 u/r) = (28.0134 u/(r + 0.000125 m))
Clearing and solving r:
r = 1.25 m = 1250 mm
