Answer:
a ship is a large vessel intended for oceangoing or at least deep-water transport, and a boat is anything else." Basically, a ship can carry a boat, but a boat cannot carry a ship
Answer:
True
Explanation:
Wastewater should never be discharged into storm sewers. Storm sewers are constructed to store rainwater.
Rainwater passes through storm sewers to the nearest underground water resource if the wastewater discharge into storm sewers, then they will also pollute the ground water, so usually the dirty water is not put in the storm sewers.
Answer:
120 Hz
Explanation:
Given that a 60 Hz sinusoidal voltage is applied to the input of a half-wave rectifier, what will be the output frequency ?
Ideally, the supply frequency will be equal to the ripple frequency.
But in a half - way rectifier, the out put frequency is twice the input sinusoidal voltage frequency.
The output frequency = 2 × 60
Output frequency = 120 Hz
Therefore, When a 60 Hz sinusoidal voltage is applied to the input of a half-wave rectifier, the output frequency is 120 Hz.
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).
