At the "very top" of the ball's path, there's a tiny instant when the ball
is changing from "going up" to "going down". At that exact tiny instant,
its vertical speed is zero.
You can't go from "rising" to "falling" without passing through "zero vertical
speed", at least for an instant. It makes sense, and it feels right, but that's
not good enough in real Math. There's a big, serious, important formal law
in Calculus that says it. I think Newton may have been the one to prove it,
and it's named for him.
By the way ... it doesn't matter what the football's launch angle was,
or how hard it was kicked, or what its speed was off the punter's toe,
or how high it went, or what color it is, or who it belongs to, or even
whether it's full to the correct regulation air pressure. Its vertical speed
is still zero at the very top of its path, as it's turning around and starting
to fall.
Explanation:
a) glowing embers in a campfire : It is a non mechanical conversion
(b) a strong wind: Kinetic or mechanical
(c) a swinging pendulum: A combination of potential and kinetic energy.
(d) a person sitting on a mattress: potential energy or mechanical
(e) a rocket being launched into space
: kinetic and potential.
Answer:
it is very large and invisible
Answer:
Where m is Ashley's mass and M is Miranda mass, in kg.
Explanation:
Let m be Ashley's mass and M be Miranda mass, in kg.
By the law of momentum conservation, after the hop on, we have the following momentum equation:
mv + MV = (m + M)S
where v = 3m/s is Ashley speed before the hop on, V = 4.2 m/s is Miranda's speed before the hop on. And S is their speed after
A) 222 N/m
Explanation:
Hook's law gives us the relationship between force (F), displacement with respect to the original length (x) and spring's constant (k):
In this part of the problem, we have:
F = 120 N - 100 N = 20 N is the new force applied
x = 9.0 cm = 0.09 m is the displacement relative to the initial stretched position
Solving the equation for k, we find the spring constant
B) 45 cm
We can use Hook's law also for this part of the exercise:
where this time we have
F = 100 N (the original pull applied)
k = 222 N/m
Solving the equation for x, we find the original displacement: