Answer:
<em>The comoving distance and the proper distance scale</em>
<em></em>
Explanation:
The comoving distance scale removes the effects of the expansion of the universe, which leaves us with a distance that does not change in time due to the expansion of space (since space is constantly expanding). The comoving distance and proper distance are defined to be equal at the present time; therefore, the ratio of proper distance to comoving distance now is 1. The scale factor is sometimes not equal to 1. The distance between masses in the universe may change due to other, local factors like the motion of a galaxy within a cluster. Finally, we note that the expansion of the Universe results in the proper distance changing, but the comoving distance is unchanged by an expanding universe.
We don't know Carter, and we don't know where he is or what
he's doing, so I'm taking a big chance speculating on an answer.
I'm going to say that if Carter is pretty much just standing there,
or, let's say, lying on the ground taking a nap, then the force of
the ground acting on him is precisely exactly equal to his weight.
Answer:
The net acceleration of the boat is approximately 6.12 m/s² downwards
Explanation:
The buoyant or lifting force applied to the boat = 790 N
The mass of the boat lifted by the buoyant force = 214 kg
The force applied to a body is defined as the product of the mass and the acceleration of the body. Therefore, the buoyant force, F, acting on the boat can be presented as follows;
Fₐ = F - W
The weight of the boat = 214 × 9.81 = 2099.34 N
Therefore;
Fₐ = 790 - 2099.34 = -1309.34 N
Fₐ = Mass of the boat × The acceleration of the boat
Given that the buoyant force, Fₐ, is the net force acting on the boat, we have;
F = Mass of the boat × The net acceleration of the boat
F = -1309.34 N = 214 kg × The net acceleration of the boat
∴ The net acceleration of the boat = -1309.34 N/(214 kg) ≈ -6.12 m/s²
The net acceleration of the boat ≈ 6.12 m/s² downwards