Answer:
a = - 50 [m/s²]
Explanation:
To solve this problem we simply have to replace the values supplied in the given equation.
Vf = final velocity = 0.5 [m/s]
Vi = initial velocity = 10 [m/s]
s = distance = 100 [m]
a = acceleration [m/s²]
Now replacing we have:
![(0.5)^{2}-(10)^{2} = 2*a*(100)\\0.25-10000=200*a\\200*a=-9999.75\\a =-50 [m/s^{2} ]](https://tex.z-dn.net/?f=%280.5%29%5E%7B2%7D-%2810%29%5E%7B2%7D%20%3D%202%2Aa%2A%28100%29%5C%5C0.25-10000%3D200%2Aa%5C%5C200%2Aa%3D-9999.75%5C%5Ca%20%3D-50%20%5Bm%2Fs%5E%7B2%7D%20%5D)
The negative sign of acceleration means that the ship slows down its velocity in order to land.
A. 0.5kg
To get this answer you need to follow the equation of KE=0.5*mv^2
But we don't have the m part in the equation. So just plug in the numbers to see which works best, though I can tell you before we do that the answer would be a.
As you may know, gravity, is a force of 9.8 m/s. And we want to get 9.8 Joules. So if we take a half a kg stone, release it at one meter, we get half of the normal gravity pull, 4.90 Joules. That means if we take half a kg stone and drop it at a doubled height, we get 9.8 Joules.
That is also to say that if we have a 1kg stone and drop it at one meter you will get the normal pull of gravity in Joules, 9.8J.
Be careful though, this does not mean if you drop a 1kg stone and a .5 kg stone the 1kg will hit first. This simply means that the 1kg stone will have twice the Joules that the .5kg stone has.
Answer: I would say the answer is B.
Explanation: It looks and sounds the most correct.
Answer:
The question is wrong Since if you apply Force on 0.0m²It would mean That the pressure exerted=F/A=F/0
An since we can't divide a number by 0, the question is wrong
The direction of the motion is constantly changing during
motion over any closed path, not only circular.
