Answer:
<em>The comoving distance and the proper distance scale</em>
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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.
The radius, r, of the child from the center of the wheel is
r = 1.3 m
The wheel makes one revolution in 4.2 s. Its angular velocity is
ω = (2π rad)/(4.2 s) = 1.496 rad/s
The linear speed of the child is the tangential velocity, given by
v = rω
= (1.3 m)*(1.496 rad/s)
= 1.945 m/s
Answer: 1.95 m/s (nearest hundredth)
Answer:
a) Tc = 750 [N] ;b) See the explanation below.
Explanation:
To solve this problem, we first need a graphical explanation of this, as well as knowing the corresponding questions. Therefore, a search was carried out in google, in the attached image we will find a graphical description of the problem.
b)
The solution of this type of problem corresponds to the use of Newton's third law, applying static which tells us that the sum of the forces in a system in equilibrium without movement must be equal to zero.
a)
In this way we can find by means of a sum of forces on the y axis equal to zero:
- 850 - 450 + 550 + Tc = 0
Tc = 750 [N]
Answer:
The floor of the ocean is 6120 m deep.
Explanation:
In order to find the depth of the ocean we need to use the speed of the ultrasonic sound 1530 m/s and the time it takes for the echo to comeback. Since the wave is transmitted by the vessel goes to the bottom of the ocean and comeback, it travels the distance between the vessel and the floor two times, so we can divide the time by two. We then have:
D = V*t/2 = 1530*8/2 = 1530*4 = 6120 m