Answer:
μ = mg/kx
Explanation:
Since the bock does not slip, the frictional force equals the weight of the block. So, F = mg. Now, the frictional force, F = μN where μ = coefficient of static friction and N = Normal force.
Now, the normal force equals the spring force F' = kx where k = spring constant and x = compression of spring.
N = F' = kx
So, F = μN = μkx
μkx = mg
So, μ = mg/kx
The centripetal force acting on the ball will be 23.26 N.The direction of the centripetal force is always in the path of the center of the course.
<h3>What is centripetal force?</h3>
The force needed to move a body in a curved way is understood as centripetal force. This is a force that can be sensed from both the fixed frame and the spinning body's frame of concern.
The given data in the problem is;
m is the mass of A ball = 0.25 kg
r is the radius of circle= 1.6 m rope
v is the tangential speed = 12.2 m/s
is the centripetal force acting on the ball
The centripetal force is found as;

Hence the centripetal force acting on the ball will be 23.26 N.
To learn more about the centripetal force refer to the link;
brainly.com/question/10596517
It has to due with numbers so I would say the last one!
Answer:
This reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen. Add 2 molecules of hydrogen peroxide and 2 molecules of water. Since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.
3]Explanation: This reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen. Add 2 molecules of hydrogen peroxide and 2 molecules of water. Since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.
4]Two moles of hydrogen peroxide H2O2 decomposes to produce two moles of water H2O and one mole of oxygen gas O2(g) , which then bubbles off
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