<span>high pressure produced by the clouds because its the most likely!!!!!!!!!!</span>
Answer:
Water can spread germs through bacteria.
Explanation:
Bacteria can live in water and can spread to new locations when water is moved.
This is a question on conservation of energy. That is,
mgh + KE1 = KE2
mgh +1/2mv1^2 = 1/2mv2^2
gh + 1/2v1^2 = 1/2v2^2
Where, h = 0.2 m, v1 =3.04 m/s
Therefore,
v2 = Sqrt [2(gh+1/2v1^2)] = Sqrt [2(9.81*0.2 + 1/2*3.04^2)] = 7.26 m/s
Now, Volumetric flow rate, V/time, t = Surface area, A*velocity, v
Where,
V = Av = πD^2/4*3.04 = π*(2.51/100)^2*1/4*3.04 = 1.504*10^-3 m^3/s
At 0.2 m below,
V = 1.504*10^-3 m^3/s = A*7.26
A = (1.504*10^-3)/7.26 = 2.072*10^-4 m^2
But, A = πr^2
Then,
r = Sqrt (A/π) = Sqrt (2.072*10^-4/π) = 0.121*10^-3 m
Diameter = 2r = 0.0162 m = 1.62 cm
Answer:
the rate of heat transfer after the system achieves steady state is -3.36 kW
Explanation:
Given the data in the question;
mass of water m = 50 kg
N = 300 rpm
Torque T = 0.1 kNm
V = 110 V
I = 2 A
Electric work supplied W₁ = PV = 2 × 110 = 220 W = 0.22 kW
Now, work supplied by paddle wheel W₂ is;
W₂ = 2πNT/60
W₂ = (2π × 0.1 × 300) / 60
W₂ = 188.495559 / 60
W₂ = 3.14 kW
So the total work will be;
W = 0.22 + 3.14
W = 3.36 kW
Hence total work done on the system is 3.36 kW.
At steady state, the properties of the system does not change so the heat transfer will be 3.36 KW.
The heat will be rejected by the system so the sign of heat will be negative.
i.e Q = -3.36 kW
Therefore, the rate of heat transfer after the system achieves steady state is -3.36 kW
