I believe it is -1.11 m/s^2. I will let you know if its correct
We divide the thin rectangular sheet in small parts of height b and length dr. All these sheets are parallel to b. The infinitesimal moment of inertia of one of these small parts is

where

Now we find the moment of inertia by integrating from

to

The moment of inertia is

(from (-a/2) to

(a/2))
Answer: d= 0.57* l
Explanation:
We need to check that before ladder slips the length of ladder the painter can climb.
So we need to satisfy the equilibrium conditions.
So for ∑Fx=0, ∑Fy=0 and ∑M=0
We have,
At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal
At the top of ladder, N₂ acting horizontal
And Between somewhere we have the weight of painter acting downward equal to= mg
So, we have N₁=mg
and also mg*d*cosФ= N₂*l*sin∅
So,
d=
* tan∅
Also, we have f₁=N₂
As f₁= чN₁
So f₁= 0.357 * 69.1 * 9.8
f₁= 241.75
Putting in d equation, we have
d=
* tan 58
d= 0.57* l
So painter can be along the 57% of length before the ladder begins to slip
Answer:
38.6 J
Explanation:
c = molar heat capacity of silver = 25.35 J/(mol °C)
m = given mass of silver = 9.00 g
M = Molar mass of silver = 108 g
n = Number of moles of silver
Number of moles of silver are given as


n = 0.0833
Q = Energy needed to raise the temperature
ΔT = Change in temperature = 18.3 °C
Energy needed to raise the temperature is given as
Q = n c ΔT
Q = (0.0833) (25.35) (18.3)
Q = 38.6 J
What is the question yes it is converted to and then released