A stone is thrown vertically upward with a speed of 17.0 m/s. How fast is it moving when it reaches a height of 11.0 m? How long is required to reach this height?
Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):
s = ut + ½ at^2 …. (1)
v^2 = u^2 + 2as …. (2)
v = u + at …. (3)
s = (u + v)t/2 …. (4)
where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.
In this case, we know u = 17.0m/s, a = -g = -9.81m/s^2, s = 11.0m and we want to know v and t, so from equation (2):
v^2 = u^2 + 2as
v^2 = 17.0^2 -2(9.81)(11.0)
v = √73.18 = 8.55m/s
now from equation (3):
v = u + at
8.55 = 17.0 – 9.81t
t = (8.55 – 17.0)/(-9.81) = 0.86s
Answer:
A
Explanation:
ewan ang hirap naman nyan
We don't know. A black hole is a star that has collapsed into its own gravity. The gravity in fact, is so strong that even light cannot get through it. That's why it looks black to us.
They have a different number neutrons.
Answer:
67000N
Explanation:
We solve for the acceleration using the the 3rd constant-acceleration equation.
(Vx)f² = (Vx)i² + 2ax∆x
We have the displacement to be
∆x = Xf - Xi = 940m
Vx = 70m/s
The acceleration = (70m/s)²/2(940m)
= 4900/1880
= 2.61m/s²
From isaac newton's second law,
51000kg x 2.61m/s²
= 133,000N
The engines thrust is half of this value
Therefore thrust = 67000N or 67kN
