The formula for the momentum is
p = mv
As a consequence of the conservation of energy, there is also the law of conservation of momentum.
So,
Δp = Δmv = mΔv
The maximum precision that the position can be ascertained is less than 100% because of dissipation.
Answer:
A. Two protons and two neutrons.
Explanation:
Alpha particles are positively charged specie identical to the nucleus of Helium atom.
Answer:
उव्ग्वुव ह्व्झ एउएइहे एइएइएइएएइ सिसुब्स्सी बीस सिस इस्ब एइब
Explanation:
?उग्व्ब्वु विब्सिए इसिग्व विद्बिअब्द सिह्व्व इस्ब्व दिव्ब्स विह्द ऐद्जिइ सुउगव्दी सिइगैगे क्ज्गैइव अजिव्व्ज्व्स कैह्द अजि ह्ज्फ्ज इअह इकुगै ईग इअबे अजिव्ब जैइअब इऐहे ऐइहे ऐइग्गे अत्व्ब ओप्झब रोज दिधिए ऊइफ्ब इसुहद ईउहे सिउउअ दिइब्द स्सिउए ऐइहे सिएय्व एउविये एइव्वे
Rigidbodies are components that allow a GameObject<u> to react to real-time physics. </u>
Explanation:
- Rigidbodies are components that allow a GameObject to react to real-time physics. This includes reactions to forces and gravity, mass, drag and momentum. You can attach a Rigidbody to your GameObject by simply clicking on Add Component and typing in Rigidbody2D in the search field.
- A rigidbody is a property, which, when added to any object, allows it to interact with a lot of fundamental physics behaviour, like forces and acceleration. You use rigidbodies on anything that you want to have mass in your game.
- You can indeed have a collider with no rigidbody. If there's no rigidbody then Unity assumes the object is static, non-moving.
- If you had a game with only two objects in it, and both move kinematically, in theory you would only need a rigidbody on one of them, even though they both move.
<span>1/3
The key thing to remember about an elastic collision is that it preserves both momentum and kinetic energy. For this problem I will assume the more massive particle has a mass of 1 and that the initial velocities are 1 and -1. The ratio of the masses will be represented by the less massive particle and will have the value "r"
The equation for kinetic energy is
E = 1/2MV^2.
So the energy for the system prior to collision is
0.5r(-1)^2 + 0.5(1)^2 = 0.5r + 0.5
The energy after the collision is
0.5rv^2
Setting the two equations equal to each other
0.5r + 0.5 = 0.5rv^2
r + 1 = rv^2
(r + 1)/r = v^2
sqrt((r + 1)/r) = v
The momentum prior to collision is
-1r + 1
Momentum after collision is
rv
Setting the equations equal to each other
rv = -1r + 1
rv +1r = 1
r(v+1) = 1
Now we have 2 equations with 2 unknowns.
sqrt((r + 1)/r) = v
r(v+1) = 1
Substitute the value v in the 2nd equation with sqrt((r+1)/r) and solve for r.
r(sqrt((r + 1)/r)+1) = 1
r*sqrt((r + 1)/r) + r = 1
r*sqrt(1+1/r) + r = 1
r*sqrt(1+1/r) = 1 - r
r^2*(1+1/r) = 1 - 2r + r^2
r^2 + r = 1 - 2r + r^2
r = 1 - 2r
3r = 1
r = 1/3
So the less massive particle is 1/3 the mass of the more massive particle.</span>
