Answer:
im pretty sure its c the third answer i got that one right
Explanation:
your welcome :)
The re<span>sistance of the second wire is 16 R.
where R is the resistance of the first wire.
R = </span>ρ
where l = length of the wire
A = area of the wire
A =
where, r =
Thus, on finding the ratio of resistance of the two wires, we get,
here, R1 = R
l1 = 8m
l2 = 2m
A1=π
A1=π
we get. R2 = 16R
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:
The component of block weight parallel to the plane, Wₓ = W cosθ
Explanation:
Let the weight of the block due to gravitation is W
The direction of the weight is vertically down
Let θ be the angle formed with the vertical weight of the block and the incline.
Taking two components of weight one along the vertical weight and another component perpendicular to it.
Then the component `of weight long the parallel of the plane is
Wₓ = W cosθ
