Answer: 12.4 feet
Explanation:
If there is a smooth transition and there is no change in slopes, energy considerations can be used
The cube has a kinetic energy of
ke = mv^2/2 = 10 lbm * 20^2ft^2/s^2 / 2 = 2000 lbm-ft^2 / s^2
At the highest point when there is a gain in potential energy
pe = mgh = 10 lbm * 32.2 ft/s^2 * h ft = 322 lbm ft^2/s^2
If there is no loss in energies,
pe = ke
322h lbm ft^2/s^2 = 2000 lbm ft^2/s^2
h = 2000 /322 = 6.211 (ft)
= h / sin(30) = 12.4 ft
Answer:
The maximum load the bar can withstand = 35.43 KN
Explanation:
Ultimate tensile strength of the given aluminium bar
= 540 M pa
Cross section area of the bar =
= 65.61 
We know that the ultimate strength of the bar is calculated from


= 540 × 65.61
= 35.43 KN
Therefore the maximum load the bar can withstand = 35.43 KN
Answer:
The statement regarding the mass rate of flow is mathematically represented as follows 
Explanation:
A junction of 3 pipes with indicated mass rates of flow is indicated in the attached figure
As a basic sense of intuition we know that the mass of the water that is in the pipe junction at any instant of time is conserved as the junction does not accumulate any mass.
The above statement can be mathematically written as

this is known as equation of conservation of mass / Equation of continuity.
Now we know that in a time 't' the volume that enter's the Junction 'O' is
1) From pipe 1 = 
1) From pipe 2 = 
Mass leaving the junction 'O' in the same time equals
From pipe 3 = 
From the basic relation of density, volume and mass we have

Using the above relations in our basic equation of continuity we obtain

Thus the mass flow rate equation becomes 
Answer
63 %
Explanation:
It is given that the reservoirs is at the temperature of 627°C and 27°C
So lower temperature that is
= 60°C=273+60=333 K
And the higher temperature that is
= 627°C =273+627=900 K
We know that the thermal efficiency of thermal reservoir = 1-
=
= 0.63 =63 %
So the efficiency of the reservoir is 63%