When you first pull back on the pendulum, and when you pull it back really high the Potential Energy is high and the Kinetic Energy is low, But when up let go, and it gets right around the middle, that's when the Potential energy transfers to Kinetic, at that point the kinetic Energy is high and the potential Energy is low. But when it comes back up at the end. The same thing will happen, the Potential Energy is high, and the Kinetic Energy is low. Through all of that the Mechanical Energy stays the same.
I hope this helps. :)
Brainliest?
The solution would be like
this for this specific problem:
<span>
The force on m is:</span>
<span>
GMm / x^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2] ->
1
The force on 2m is:</span>
<span>
GM(2m) / (L - x)^2 + Gm(2m) / L^2 = 2[Gm (2m) / L^2]
-> 2
From (1), you’ll get M = 2mx^2 / L^2 and from
(2) you get M = m(L - x)^2 / L^2
Since the Ms are the same, then
2mx^2 / L^2 = m(L - x)^2 / L^2
2x^2 = (L - x)^2
xsqrt2 = L - x
x(1 + sqrt2) = L
x = L / (sqrt2 + 1) From here, we rationalize.
x = L(sqrt2 - 1) / (sqrt2 + 1)(sqrt2 - 1)
x = L(sqrt2 - 1) / (2 - 1)
x = L(sqrt2 - 1) </span>
= 0.414L
<span>Therefore, the third particle should be located the 0.414L x
axis so that the magnitude of the gravitational force on both particle 1 and
particle 2 doubles.</span>
Answer:
option C
Explanation:
The correct answer is option C
Fire cut of fireman cut is diagonal cut which is provided at the end of the beam to prevent the fall of masonry wall if a fire breaks out in the building.
Fire cut allows joist to leave if it fails without affecting the masonry wall standing.
Without fire cut, the burnt beam will rotate downward affecting the connection of beam and wall and leading to damage it.
This problem is a piece o' cake, IF you know the formulas for both kinetic energy and momentum. So here they are:
Kinetic energy = (1/2) · (mass) · (speed²)
Momentum = (mass) · (speed)
So, now ... We know that
==> mass = 15 kg, and
==> kinetic energy = 30 Joules
Take those pieces of info and pluggum into the formula for kinetic energy:
Kinetic energy = (1/2) · (mass) · (speed²)
30 Joules = (1/2) · (15 kg) · (speed²)
60 Joules = (15 kg) · (speed²)
4 m²/s² = speed²
Speed = 2 m/s
THAT's all you need ! Now you can find momentum:
Momentum = (mass) · (speed)
Momentum = (15 kg) · (2 m/s)
<em>Momentum = 30 kg·m/s</em>
<em>(Notice that in this problem, although their units are different, the magnitude of the KE is equal to the magnitude of the momentum. When I saw this, I wondered whether that's always true. So I did a little more work, and I found out that it isn't ... it's a coincidence that's true for this problem and some others, but it's usually not true.)</em>
