It might be 144.2 m but i’m not for sure:)
........................................................................................
Answer:
τ = 32.8635 N-m (counterclockwise)
Explanation:
Given
M = 2 kg
L = 2 m
r = 0.10 m
m₁ = 4 kg
r₁ = (1.00-0.30)m = 0.70 m
m₂ = 5 Kg
r₂ = (0.90-0.75)m = 0.15 m
In order to determine the torque on the meterstick if it is received and allowed to pivot about the 90 cm mark (ypu can see the pic to understand the question), we apply:
τ = r₁*m₁*g + r₂*m₂*g - r*M*g = g*(r₁*m₁+r₂*m₂-r*M)
⇒ τ = (9.81 m/s²)(0.70 m*4 kg + 0.15 m*5 Kg - 0.10 m*2 kg)
⇒ τ = 32.8635 N-m (counterclockwise)
Since the ladder is standing, we know that the coefficient
of friction is at least something. This [gotta be at least this] friction
coefficient can be calculated. As the man begins to climb the ladder, the
friction can even be less than the free-standing friction coefficient. However,
as the man climbs the ladder, more and more friction is required. Since he
eventually slips, we know that friction is less than what's required at the top
of the ladder.
The only "answer" to this problem is putting lower
and upper bounds on the coefficient. For the lower one, find how much friction
the ladder needs to stand by itself. For the most that friction could be, find
what friction is when the man reaches the top of the ladder.
Ff = uN1
Fx = 0 = Ff + N2
Fy = 0 = N1 – 400 – 864
N1 = 1264 N
Torque balance
T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)
N2 = 439 N
Ff = 439= u N1
U = 440 / 1264 = 0.3481
Answer: A. Radiation can move through empty space to transfer heat; conduction cannot.
Now unlike convection and conduction, radiation does not require matter to transfer heat. It also does not require a medium to transfer heat, crossing off the other options regardless.
I hope this helps!
