Ke= 1/2 mv ^ 2
15x 15 = 225
225 x 30 = 6750
6750/2 = 3375 J
Ke = 3375 J
–9.8 m/s<span>2
Just took it and got it right!</span>
Assuming constant acceleration <em>a</em>, the object has undergoes an acceleration of
<em>a</em> = (50 m/s - 100 m/s) / (25 s) = -2 m/s²
Then the net force has a magnitude <em>F</em> such that, by Newton's second law,
<em>F</em> = (75.0 kg) <em>a</em>
<em>F</em> = (75.0 kg) (-2 m/s²)
<em>F</em> = -150 N
meaning the object is acted upon by a net force of 150 N in the direction opposite the initial direction in which the object is moving.
Answer:
The Roche limit for the Moon orbiting the Earth is 2.86 times radius of Earth
Explanation:
The nearest distance between the planet and its satellite at where the planets gravitational pull does not torn apart the planets satellite is known as Roche limit.
The relation to determine Roche limit is:
![Roche\ limit=2.423\times R_{P}\times\sqrt[3]{\frac{D_{P} }{D_{M} } }](https://tex.z-dn.net/?f=Roche%5C%20limit%3D2.423%5Ctimes%20R_%7BP%7D%5Ctimes%5Csqrt%5B3%5D%7B%5Cfrac%7BD_%7BP%7D%20%7D%7BD_%7BM%7D%20%7D%20%7D)
....(1)
Here 
is radius of planet and 
are density of planet and moon respectively.
According to the problem,
Density of Earth,
= 5.5 g/cm³
Density of Moon,
= 3.34 g/cm³
Consider 
be the radius of the Earth.
Substitute the suitable values in the equation (1).
![Roche\ limit=2.423\times R_{E}\times\sqrt[3]{\frac{5.5 }{3.34 } }](https://tex.z-dn.net/?f=Roche%5C%20limit%3D2.423%5Ctimes%20R_%7BE%7D%5Ctimes%5Csqrt%5B3%5D%7B%5Cfrac%7B5.5%20%7D%7B3.34%20%7D%20%7D)
I would think it would be the same if you are weighting the dish and the Ice cube at the same time anyway. Not to sure though I'm a beginner and math is complicated for me sometimes.
