If they become closer, it is increased, and if the objects become farther away is decreased.
Speed is v = d/t
Or speed is distance over time
So...
40min / 60min = 0.6667 or 2/3 --> Finding what proportion 40 minutes is to an hour or 60 minutes as we need the units of hours to match up
45km/h = d/0.6667h
d = (45)(0.667)
d = 30.0015 or 30km
Asterism, a pattern of stars that is not a constellation. An asterism can be part of a constellation, such as the Big Dipper, which is in the constellation Ursa Major, and can even span across constellations, such as the Summer Triangle, which is formed by the three bright stars Deneb, Altair, and Vega.
If the three spoon touch nothing happens because they are all at room Temperature
To solve this problem we will apply the energy conservation theorem for which the work applied on a body must be equivalent to the kinetic energy of this (or vice versa) therefore


Here,
m = mass
= Velocity (Final and initial)
First case) When the particle goes from 10m/s to 20m/s



Second case) When the particle goes from 20m/s to 30m/s



As the mass of the particle is the same, we conclude that more energy is required in the second case than in the first, therefore the correct answer is A.