The number of years that will pass before the radius of the Moon's orbit increases by 3.6 x 10^6 m will be 90000000 years.
<h3>How to compute the value?</h3>
From the information given, the orbit of the moon is increasing in radius at approximately 4.0cm/yr.
Therefore, we will convey the centimeters to meter. This will be 4cm will be:
= 4/100 = 0.04m/yr.
Time = Distance / Speed
Time = 3.6 x 10^6/0.04
Time = 90000000 years.
Learn more about moon on:
brainly.com/question/13262798
#SPJ1
Complete question:
Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the moon is increasing in radius at approximately 4.0cm/y. Assuming this rate to be constant how many years will pass before the radius of the Moon's orbit increases by 3.6 x 10^6
Answer:
Explanation:
Normal length of spring = 28.3 cm
stretched length of spring = 38.2 cm
length of extension = 38.2 - 28.3 = 9.9 cm
= 9.9 x 10⁻² m
force applied to stretch = .55 x 9.8 ( mg )
= 5.39 N
Force constant = force applied / extension
= 5.39 / 9.9 x 10⁻²
= .5444 x 10² N /m
= 54.44 N/m
I'd go for b) v squared.
Wind is air in motion. So, it has kinetic energy. For a given mass of air at a certain speed, the kinetic energy would be (1/2)mv^2.
Since everything else in the chain seems to be proportional, then it's the v^2 bit that seems to be important. Hence the answer I post here.
MRI-Magnetic resonsance imaging
Look at the pic!! Hope
It helps love ❤️
Technically, ALL of the planets do.
The effect is greatest inside Mercury, because Mercury is the one closest to the Sun.
