The static friction exerted on the block by the incline is 
.
The given parameters;
- <em>mass of the block, = M</em>
- <em>coefficient of static friction in section 1, = </em>
<em /> - <em>angle of inclination of the plane, = θ</em>
<em />
The normal force on the block is calculated as follows;
Fₙ = Mgcosθ
The static friction exerted on the block by the incline is calculated as follows;
Thus, the static friction exerted on the block by the incline is
Learn more here:brainly.com/question/17237604
Force, newtons 3rd law of motion stated for every action there is an equal and opposite reaction
<span>With a half-life of 700 million years, U-235 would have had twice as much mass at a time 700 MYA. This would have put the mass at 100kg at that time. Going back another 50 million years would be (50/700) or 1/14 of the half-life, or (1/2 * 1/14), or 1/28 of the total mass. 1/28 of 100kg is 3.57kg, so the amount present at the 750MYA mark would be approximately 103.57kg of U-235.</span>
Answer:
The force does the ceiling exert on the hook is 269.59 N
Explanation:
Applying the second Newton law:
F = m*a
From the attached diagram, the net force in object 1 is:
In object 2:
Adding the two equations:
(eq. 1)
The torque:
Where
I = moment of inertia
α = angular acceleration
If the linear acceleration is
Torque due the tension is equal:
Substituting torque, mass, in equation 1, the expression respect the acceleration is:
Where
W₁ = 75 N
W₂ = 125 N
W = 80 N
The net force is:
