Answer:
Explanation:
For this interesting problem, we use the definition of centripetal acceleration
a = v² / r
angular and linear velocity are related
v = w r
we substitute
a = w² r
the rectangular body rotates at an angular velocity w
We locate the points, unfortunately the diagram is not shown. In this case we have the axis of rotation in a corner, called O, in one of the adjacent corners we call it A and the opposite corner A
the distance OB = L₂
the distance AB = L₁
the sides of the rectangle
It is indicated that the acceleration in in A and B are related
we substitute the value of the acceleration
w² r_A = n r_B
the distance from the each corner is
r_B = L₂
r_A = 
we substitute
\sqrt{L_1^2 + L_2^2} = n L₂
L₁² + L₂² = n² L₂²
L₁² = (n²-1) L₂²
Answer:
As collision is elastic,thus we can use conservation of momentum equation
mA=0.2 kg
(vB)1=0 m/s.......................as it is on rest before collision
(vA)1=4 m/s
(vA)2=-1 m/s
(vB)2=2 m/s
using equation
(mA*vA+mB*vB)1= (mA*vA+mB*vB)2
Where 1 and 2 represents before and after collision
(0.2*4)+(mB*0)=(0.2*-1)+(mB*2)
0.8=-0.2+(2mB)
mass of object B=mB=0.3 Kg
The car has an initial velocity 
of 23 m/s and a final velocity 
of 0 m/s. Recall that for constant acceleration,
The car stops in 7 s, so the acceleration is
Answer:
x ’= 1,735 m, measured from the far left
Explanation:
For the system to be in equilibrium, the law of rotational equilibrium must be fulfilled.
Let's fix a reference system located at the point of rotation and that the anticlockwise rotations have been positive
They tell us that we have a mass (m1) on the left side and another mass (M2) on the right side,
the mass that is at the left end x = 1.2 m measured from the pivot point, the mass of the right side is at a distance x and the weight of the body that is located at the geometric center of the bar
x_{cm} = 1.2 -1
x_ {cm} = 0.2 m
Σ τ = 0
w₁ 1.2 + mg 0.2 - W₂ x = 0
x = 
x = 
let's calculate
x = 
2.9 1.2 + 4 0.2 / 8
x = 0.535 m
measured from the pivot point
measured from the far left is
x’= 1,2 + x
x'= 1.2 + 0.535
x ’= 1,735 m
You are running at constant velocity in the x direction, and based on the 2D definition of projectile motion, Vx=Vxo. In other words, your velocity in the x direction is equal to the starting velocity in the x direction. Let's say the total distance in the x direction that you run to catch your own ball is D (assuming you have actual values for Vx and D). You can then use the range equation, D= (2VoxVoy)/g, to find the initial y velocity, Voy. g is gravitational acceleration, -9.8m/s^2. Now you know how far to run (D), where you will catch the ball (xo+D), and the initial x and y velocities you should be throwing the ball at, but to find the initial velocity vector itself (x and y are only the components), you use the pythagorean theorem to solve for the hypotenuse. Because you know all three sides of the triangle, you can also solve for the angle you should throw the ball at, as that is simply arctan(y/x).
