Answer:
A) Sump pit
Explanation:
A wastewater typically refers to a body of water that has contaminated through human use in homes, offices, schools, businesses etc. Wastewater are meant to be disposed in accordance with the local regulations and standards because they are unhygienic for human consumption or use.
Generally, many homes use a floor drain in their bathrooms and toilets to remove wastewater in order to mitigate stagnation and to improve hygiene. A floor drain can be defined as a material installed on floors for the continuous removal of any stagnant wastewater in buildings. Wastewater flows into a sump pit once it is released into a floor drain through the use of a pipe such as a polyvinyl chloride (PVC) pipe, which directly connects the floor drain to the sump pit. The wastewater can the be removed from the sump pit when it is filled up through the use of a pump.
Answer:
5.833
Explanation:
Coefficient of Perfomance (COP) is the ratio of refrigeration effect to power input.
where RE is refrigeration effect and P is power input
Here, the power input is given as 30 kW
We also know that 1 ton cooling is equivalent to 3.5 kW hence for 50 tons, RE=50*3.5=175 kW
Now the 
Answer:
Absolute pressure=70.72 KPa
Explanation:
Given that Vacuum gauge pressure= 30 KPa
Barometer reading =755 mm Hg
We know that barometer always reads atmospheric pressure at given situation.So atmospheric pressure is equal to 755 mm Hg.
We know that P= ρ g h
Density of 
So P=13600 x 9.81 x 0.755
P=100.72 KPa
We know that
Absolute pressure=atmospheric pressure + gauge pressure
But here given that 30 KPa is a Vacuum pressure ,so we will take it as negative.
Absolute pressure=atmospheric pressure + gauge pressure
Absolute pressure=100.72 - 30 KPa
So
Absolute pressure=70.72 KPa
Answer:
a) the inductance of the coil is 6 mH
b) the emf generated in the coil is 18 mV
Explanation:
Given the data in the question;
N = 570 turns
diameter of tube d = 8.10 cm = 0.081 m
length of the wire-wrapped portion l = 35.0 cm = 0.35 m
a) the inductance of the coil (in mH)
inductance of solenoid
L = N²μA / l
A = πd²/4
so
L = N²μ(πd²/4) / l
L = N²μ(πd²) / 4l
we know that μ = 4π × 10⁻⁷ TmA⁻¹
we substitute
L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)
L = 0.00841549 / 1.4
L = 6 × 10⁻³ H
L = 6 × 10⁻³ × 1000 mH
L = 6 mH
Therefore, the inductance of the coil is 6 mH
b)
Emf ( ∈ ) = L di/dt
given that; di/dt = 3.00 A/sec
{∴ di = 3 - 0 = 3 and dt = 1 sec}
Emf ( ∈ ) = L di/dt
we substitute
⇒ 6 × 10⁻³ ( 3/1 )
= 18 × 10⁻³ V
= 18 × 10⁻³ × 1000
= 18 mV
Therefore, the emf generated in the coil is 18 mV
