Factor out 8 and then facotr and u get
8/9(9x+1)(9x-1
Answer:
9.9 m/s
Explanation:
t = Time taken
u = Initial velocity
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
If the body has started from rest then the initial velocity is 0. In order to find the velocity just before hitting the water then the distance at which the downward motion stops is irrelevant.
Hence, the speed of the diver just before striking the water is 9.9 m/s
This is another time to look at Newton's 2nd law of motion:
Net Force = (mass) x (acceleration)
If the object is not moving, then its acceleration is certainly zero, and Newton's law looks like this:
Net Force = (mass) x (zero)
or Net Force = (zero) .
"Net Force = zero" means that if there ARE any forces acting on the object, then they add up to zero, and we call them "balanced" forces.
So the answer is '<em>yes</em>', and that's why.
Answer:
The force becomes 16 times what it is now.
Explanation:
The formula for gravitational force is
F = G * m1 * m2 / r^2
When you do what you have described, you are setting a stage that not even the USS Enterprise (Star Trek) can get out of. The increase is huge.
If you double m1 and m2 and don't do anything to r, you've already increased the force by 4 times. (2m1 * 2m2 = 4 * m1 * m2)
But you are not finished. If you 1/2 the distance, you are again increasing the Force by 4 times. 1 / (2r) ^2 = 1/ 4* r^2
Because this is in the denominator, the 1/4 is going to flip to the numerator.
So the total increase is going to be 4 * (4 * m1 * m2) = 16 * m1 * m2.
Think about what that means. If you were out golfing, your drives would be roughly 1/16 times as far as they are now. Also you would be lugging around 16 times your weight around the golf course. My feeling is that you would never finish 5 holes at that rate.
