Answer:
57300 N
Explanation:
The container has a mass of 5300 kg, the weight of the container is:
f = m * a
w = m * g
w = 5300 * 9.81 = 52000 N
However this container was moving with more acceleration, so dynamic loads appear.
w' = m * (g + a)
w' = 5300 * (9.81 + 1) = 57300 N
The rating for the cable was 50000 N
The maximum load was exceeded by:
57300 / 50000 - 1 = 14.6%
When the sound wave returns to the machine, you can measure
how long it took to return.
(You may notice that it's working just like RADAR, which does the
same thing with radio waves instead of sound waves.)
Even if you know how long the sound took to get to the bottom and
return to the top, you can't DO anything with this information if you
don't know the SPEED of the sound through the water. Not only
the inventory of this machine, but anyone who uses it, has to know
the speed of the sound through water in order to use the round-trip
time to calculate the depth.
Answer:
The thermal conductivity of the wall = 40W/m.C
h = 10 W/m^2.C
Explanation:
The heat conduction equation is given by:
d^2T/ dx^2 + egen/ K = 0
The thermal conductivity of the wall can be calculated using:
K = egen/ 2a = 800/2×10
K = 800/20 = 40W/m.C
Applying energy balance at the wall surface
"qL = "qconv
-K = (dT/dx)L = h (TL - Tinfinity)
The convention heat transfer coefficient will be:
h = -k × (-2aL)/ (TL - Tinfinty)
h = ( 2× 40 × 10 × 0.05) / (30-26)
h = 40/4 = 10W/m^2.C
From the given temperature distribution
t(x) = 10 (L^2-X^2) + 30 = 30°
T(L) = ( L^2- L^2) + 30 = 30°
dT/ dx = -2aL
d^2T/ dx^2 = - 2a
