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.
Answer:
5 kg
Explanation:
Acceleration = 6 m/s^2
Force = 30 N
Force = mass * acceleration
mass = force / acceleration
mass = 30 / 6
mass = 5 kg
Answer: For ideal machine efficiency = 1. Hence M.A = V. R. The V. R of an ideal machine and the practical machine is a constant or is the same for both
Answer. Second Option: .85p_o=p_o e^-.00012h
Solution:
P(h)=Po e^(-0.00012h)
Air pressure: P(h)
Height above the surface of the Earth (in meters): h
Air pressure at the sea level: Po
Height at which air pressure is 85% of the air pressure at sea level:
h=?, P(h)=85% Po
P(h)=(85/100) Po
P(h)=0.85 Po
Replacing P(h) by 0.85 Po in the formula above:
P(h)=Po e^(-0.00012h)
0.85 Po = Po e^(-0.00012h)
It's hard to tell exactly what's happening in that 110 cm that you marked over the wave. What is under the ends of the long arrow ? How many complete waves ? I counted 4.5 complete waves ... maybe ?
If there are 4.5 complete waves in 110cm, then the length of 1 wave is (110/4.5)=24.44cm.
Frequency = speed/wavelength
Frequency = 2m/s /0.2444m
Frequency = 8.18 Hz
