Answer: uve earned 5 b point for helping us
Explanation:well know I'm not helping u
Answer:
ρ=962.16kg/m^3
Explanation:
The first thing we must do to solve is to find the mass of the specimen using the weight equation
w = mg
m=w/g
m=0.45/9.81=0.04587kg
To find the volume we must make a free-body diagram on the specimen, taking into account that the weight will go down and the buoyant force up, and the result of that subtraction will be the measured weight value (0.081N).
We must bear in mind that the principle of archimedes indicates that the buoyant force is given by
F = ρgV
where V is the specimen volume and ρ is the density of alcohol = 789kg / m ^ 3
considering the above we have the following equation
0.081=0.45-(789)(9.81m/s^2)V
solving for V
V=(0.081-0.45)/(-789x9.81)
V=4.7673x10^-5m^3
finally we found the density
ρ=m/v
ρ=0.04587kg/4.7673x10^-5m^3
ρ=962.16kg/m^3
Answer:
P2 = 3.9 MPa
Explanation:
Given that
T₁ = 290 K
P₁ = 95 KPa
Power P = 5.5 KW
mass flow rate = 0.01 kg/s
solution
with the help of table A5
here air specific heat and adiabatic exponent is
Cp = 1.004 kJ/kg K
and k = 1.4
so
work rate will be
W = m × Cp × (T2 - T1) ..........................1
here T2 = W ÷ ( m × Cp) + T1
so T2 = 5.5 ÷ ( 0.001 × 1.004 ) + 290
T2 = 838 k
so final pressure will be here
P2 = P1 × 
..............2
P2 = 95 × 
P2 = 3.9 MPa
Answer:
Given:
efficiency of the turbine, 
= 65% = 0.65
available gross head, 
= 45 m
head loss, 
= 5 m
Discharge, Q = 
Solution:
The nozzle is 100% (say)
Available power at the inlet of the turbine, 
is given by:
(1)
where
= density of water = 997 
acceleration due to gravity, g = 
Using eqn (1):
Also, efficency, is given by:
Answer:
Answer: (a) = 3.8187m/s, (b) =24.0858m/s (c) = = 3220.071m/s
Explanation:
du/u² = dt = ∫du/2.3183 = ∫dt
0.4313 u = t + c
(a) t = 0, u= 15m/s, c = 0.647
u = t+c/0.4313 = t + 0.647/0.4313
(a) when t= 1 u = 1+ 0.647/0.4313 = 3.8187m/s
(b) when t= 10 u = 10 + 0.647/0.4313 = 24.0858m/s
(c)when t= 1000 u = 1000 + 0.647/0.4313 = 3220.071m/s
