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:
drag coefficient for the parachute is Cd = 1.84177
Explanation:
given data
diameter = 8.0 m
average vertical speed = 3 m/s
total weight of load = 200 N
to find out
drag coefficient for the parachute
solution
we will apply here drag coefficient formula that is express as here
drag coefficient Cd = 
......................1
here w is total load and A is area and v is speed
and here ρ = 1.22 kg/cu
put here value we get
Cd = 
Cd = 1.84177
drag coefficient for the parachute is Cd = 1.84177
<u>No, since the volume of the passenger's luggage ( 45,080 cm³) exceeds the allotted volume for carry-on luggages (40,000 cm³).</u>
Explanation:
<h2>Given:</h2>
Acceptable volume = 40,000 cm³
Area of luggage = 1,960 cm²
Height of luggage = 23 cm
<h2>Question:</h2>
Is the passenger's luggeage ok to carry onto the airplane
<h2>Equation:</h2>
V = l x w x h
or we can use
V = A x h
Since A = l x w
where: V - volume
A - area
l - length
w - width
h - height
<h2>Solution:</h2>
V = A x h
V = ( 1,960 cm²)(23 cm)
V = 45,080 cm³
45,080 cm³ is greater than the acceptable volume 40,000 cm³
<h2>Final Answer:</h2><h3><u>No, since the volume of the passenger's luggage ( 45,080 cm³) exceeds the allotted volume for carry-on luggages (40,000 cm³).</u></h3><h3 />
Answer:
Computation of the load is not possible because E(test) >E(yield)
Explanation:
We are asked to ascertain whether or not it is possible to compute, for brass, the magnitude of the load necessary to produce an elongation of 7.0 mm (0.28 in.). It is first necessary/ important to compute the strain at yielding from the yield strength and the elastic modulus, and then the strain experienced by the test specimen. Then, if
E(test) is less than E(yield), deformation is elastic and the load may be computed. However is E(test) is greater than E(yield) computation/determination of the load is not possible even though defamation is plastic and we have neither a stress-strain plot or a mathematical relating plastic stress and strain. Therefore, we can compute these two values as:
Calculation of E(test is as follows)
E(test) = change in l/lo= Elongation produced/stressed tension= 7.0mm/267mm
=0.0262
Computation of E(yield) is given below:
E(yield) = σy/E=275Mpa/103 ×10^6Mpa= 0.0027
Therefore, we won't be able to compute the load because for computation to take place, E(test) <E(yield). In this case, E(test) is greater than E(yield).
Answer:
The given statement "A negative normal strain can be considered to increase or decrease volume depending on coordinate system used" is
b) False
Explanation:
Normal strain refers to the strain due to normal stress which is when the applied stress is perpendicular to the surface.
Negative normal strain results in compression or contraction further leading to a decrease in volume while a positive normal strain results in elongation thus giving rise to an increase in volume.
