Answer:
Explanation:
Given that,
Height of the bridge is 20m
Initial before he throws the rock
The height is hi = 20 m
Then, final height hitting the water
hf = 0 m
Initial speed the rock is throw
Vi = 15m/s
The final speed at which the rock hits the water
Vf = 24.8 m/s
Using conservation of energy given by the question hint
Ki + Ui = Kf + Uf
Where
Ki is initial kinetic energy
Ui is initial potential energy
Kf is final kinetic energy
Uf is final potential energy
Then,
Ki + Ui = Kf + Uf
Where
Ei = Ki + Ui
Where Ei is initial energy
Ei = ½mVi² + m•g•hi
Ei = ½m × 15² + m × 9.8 × 20
Ei = 112.5m + 196m
Ei = 308.5m J
Now,
Ef = Kf + Uf
Ef = ½mVf² + m•g•hf
Ef = ½m × 24.8² + m × 9.8 × 0
Ef = 307.52m + 0
Ef = 307.52m J
Since Ef ≈ Ei, then the rock thrown from the tip of a bridge is independent of the direction of throw
Answer:
18.5 m/s
Explanation:
On a horizontal curve, the frictional force provides the centripetal force that keeps the car in circular motion:

where
is the coefficient of static friction between the tires and the road
m is the mass of the car
g is the gravitational acceleration
v is the speed of the car
r is the radius of the curve
Re-arranging the equation,

And by substituting the data of the problem, we find the speed at which the car begins to skid:

Answer:
Hello your question is incomplete attached below is the complete question
Answer : x ( acceleration of mass 4m ) = 
The top pulley rotates because it has to keep the center of mass of the system at equilibrium
Explanation:
Given data:
mass suspended = 4 meters
mass suspended at other end = 3 meters
first we have to express the kinetic and potential energy equations
The general kinetic energy of the system can be written as
T = 
T =
also the general potential energy can be expressed as
U = 
The Lagrangian of the problem can now be setup as

next we will take the Euler-Lagrange equation for the generalized equations :
Euler-Lagrange equation = 
solving the equations simultaneously
x ( acceleration of mass 4m ) = 
The top pulley rotates because it has to keep the center of mass of the system at equilibrium