Hi! I believe the answer is 2 meters(:
Answer:
(a) the velocity ratio of the machine (V.R) = 1
(b) The mechanical advantage of the machine (M.A) = 0.833
(c) The efficiency of the machine (E) = 83.3 %
Explanation:
Given;
load lifted by the pulley, L = 400 N
effort applied in lifting the, E = 480 N
distance moved by the effort, d = 5 m
(a) the velocity ratio of the machine (V.R);
since the effort applied moved downwards through a distance of d, the load will also move upwards through an equal distance 'd'.
V.R = distance moved by effort / distance moved by the load
V.R = 5/5 = 1
(b) The mechanical advantage of the machine (M.A);
M.A = L/E
M.A = 400 / 480
M.A = 0.833
(c) The efficiency of the machine (E);

Answer:
(b)False
Explanation:
defined as
=
Where x is the distance from centroidal x-axis
y is the distance from centroidal y-axis
dA is the elemental area.
The product of x and y can be positive or negative ,so the value of
can be positive as well as negative .
So from the above expressions we can say that the product of
is different from
.
Answer:
Now find the temperature of each surface, we have that the the temperature on the left side of the wall is T∞₁ - Q/h₁A and the temperature on the right side of the wall is T∞₂ + Q/h₂A.
Note: kindly find an attached diagram to the complete question given below.
Sources: The diagram/image was researched and taken from Slader website.
Explanation:
Solution
Let us consider the rate of heat transfer through the plane wall which can be obtained from the relations given below:
Q = T∞₁ -T₁/1/h₁A = T₁ -T₂/L/kA =T₂ -T∞₂/1/h₂A
= T∞₁ - T∞₂/1/h₁A + L/kA + 1/h₂A
Here
The convective heat transfer coefficient on the left side of the wall is h₁, while the convective heat transfer coefficient on the right side of the wall is h₂. the thickness of the wall is L, the thermal conductivity of the wall material is k, and the heat transfer area on one side of the wall is A. Q is refereed to as heat transfer.
Thus
Let us consider the convection heat transfer on the left side of the wall which is given below:
Q = T∞₁ -T₁/1/h₁A
T₁ = T∞₁ - Q/h₁A
Therefore the temperature on the left side of the wall is T∞₁ - Q/h₁A
Now
Let us consider the convection heat transfer on the left side of the wall which is given below:
Q= T₂ -T∞₂/1/h₂A
T₂ = T∞₂ + Q/h₂A
Therefore the temperature on the right side of the wall is T∞₂ + Q/h₂A