A boy weighing 108-lb starts from rest at the bottom A of a 6-percent incline and increases his speed at a constant rate to 7 mi
1 answer:
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Answer:
Relative density = 0.545
Degree of saturation = 24.77%
Explanation:
Data provided in the question:
Water content, w = 5%
Bulk unit weight = 18.0 kN/m³
Void ratio in the densest state, = 0.51
Void ratio in the loosest state, = 0.87
Now,
Dry density,
= 17.14 kN/m³
Also,
here, G = Specific gravity = 2.7 for sand
or
e = 0.545
Relative density =
=
= 0.902
Also,
Se = wG
here,
S is the degree of saturation
therefore,
S(0.545) = (0.05)()2.7
or
S = 0.2477
or
S = 0.2477 × 100% = 24.77%
Answer:
BOD5 = 200 mg/L
Explanation:
given data
diluted = 1% = 0.01
time = 5 day
oxygen consumption = 2.00 mg · L−1
solution
we get here BOD5 that is BOD after 5 day
and here total volume is 100% = 1
so dilution factor is = 100
so BOD5 is
BOD5 = oxygen consumption × dilution factor
BOD5 = 2 × 100
BOD5 = 200 mg/L
Answer:
Rate of heat transfer to the room air per meter of pipe length equals 521.99 W/m
Explanation:
Since it is given that the radiation losses from the pipe are negligible thus the only mode of heat transfer will be by convection.
We know that heat transfer by convection is given by
where,
h = heat transfer coefficient = 10.45 (free convection in air)
A = Surface Area of the pipe
Applying the given values in the above formula we get