Complete Question
The complete question is shown on the first uploaded image
Answer:
a
The torque produced by the pile of rocks is 
b
The distance of the single for equilibrium to occur is 
Explanation:
From the question we are told that
The mass of the left rock is 
The mass of the rock on the right 
The distance from fulcrum to the center of the pile of rocks is 
Generally the torque produced by the pile of rock is mathematically represented as
Substituting values
Generally we can mathematically evaluated the distance of the the single rock that would put the system in equilibrium as follows
The torque due to the single rock is
At equilibrium the both torque are equal
Making 
the subject of the formula
Substituting values
Answer:
cross out the false piece in blue and write the true piece in red
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.
Answer:
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Explanation:
The additional information to the question is embedded in the diagram attached below:
The height between the dragster and ground is considered to be 0.35 m since is not given ; thus in addition win 0.75 m between the dragster and the parachute; we have: (0.75 + 0.35) m = 1.1 m
Balancing the equilibrium about point A;
F(1.1) - mg (1.25) = 
- 1200(9.8)(1.25) = 1200a(0.35)
- 14700 = 420 a ------- equation (1)
--------- equation (2)
Replacing equation 2 into equation 1 ; we have :
1320 a - 14700 = 420 a
1320 a - 420 a =14700
900 a = 14700
a = 14700/900
a = 16.33 m/s²
The deceleration of the dragster upon releasing the parachute such that the wheels at B are on the verge of leaving the ground is 16.33 m/s²
Answer:
x = 76.5 m
Explanation:
Let's use Newton's second law at the point of contact between the wheel and the floor.
fr = m a
fr = miy N
N-W = 0
N = W
μ mg = m a
a = miu g
a = 0.600 9.8
a = 5.88 m / s²
Having the acceleration we can use the kinematic relationships to find the distance
² = v₀² + 2 a x
= 0
x = -v₀² / 2 a
Acceleration opposes the movement by which negative
x = - 30²/2 (-5.88)
x = 76.5 m
