<span>Because P = W ÷ t, and W = F*t, you can substitute (W) for (F*t). Then substitute (F) for (m*a). This will leave you with P = (m*a*d)/t. Since you need velocity, youd want to solve for a so you can use v = a*t. a = (P*t)/(m*d) therefore, substituting a in v = a*t, v = (P*t*t)/(m*d)</span>
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.
The answer is the third graph
Answer:
(5g/cm³)*(10cm³) = 50g
Explanation:
This is just a conversion formula. Easy to find using dimensional analysis.
(5g/cm³)*(10cm³) = 50g
Answer:
0.572
Explanation:
First examine the force of friction at the slipping point where Ff = µsFN = µsmg.
the mass of the car is unknown,
The only force on the car that is not completely in the vertical direction is friction, so let us consider the sums of forces in the tangential and centerward directions.
First the tangential direction
∑Ft =Fft =mat
And then in the centerward direction ∑Fc =Ffc =mac =mv²t/r
Going back to our constant acceleration equations we see that v²t = v²ti +2at∆x = 2at πr/2
So going backwards and plugging in Ffc =m2atπr/ 2r =πmat
Ff = √(F2ft +F2fc)= matp √(1+π²)
µs = Ff /mg = at /g √(1+π²)=
1.70m/s/2 9.80 m/s² x√(1+π²)= 0.572
