Answer: µ=0.205
Explanation:
The horizontal forces acting on the ladder are the friction(f) at the floor and the normal force (Fw) at the wall. For horizontal equilibrium,
f=Fw
The sum of the moments about the base of the ladder Is 0
ΣM = 0 = Fw*L*sin74.3º - (25.8kg*(L/2) + 67.08kg*0.82L)*cos74.3º*9.8m/s²
Note that it doesn't matter WHAT the length of the ladder is -- it cancels.
Solve this for Fw.
0= 0.9637FwL - (67.91L)2.652
Fw=180.1/0.9637
Fw=186.87N
f=186.81N
Since Fw=f
We know Fw, so we know f.
But f = µ*Fn
where Fn is the normal force at the floor --
Fn = (25.8 + 67.08)kg * 9.8m/s² =
910.22N
so
µ = f / Fn
186.81/910.22
µ= 0.205
Answer is B.
In a lever, the effort arm is 2 times as a long as the load arm. The resultant force will be twice the applied force.
Hope it helped you.
-Charlie
Answer : The temperature when the water and pan reach thermal equilibrium short time later is, 
Explanation :
In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.


where,
= specific heat of aluminium = 
= specific heat of water = 
= mass of aluminum = 0.500 kg = 500 g
= mass of water = 0.250 kg = 250 g
= final temperature of mixture = ?
= initial temperature of aluminum = 
= initial temperature of water = 
Now put all the given values in the above formula, we get:


Therefore, the temperature when the water and pan reach thermal equilibrium short time later is, 
The correct answer is: C. It is the point beyond which neither light nor anything else can escape.