Specific Gravity of the fluid = 1.25
Height h = 28 in
Atmospheric Pressure = 12.7 psia
Density of water = 62.4 lbm/ft^3 at 32F
Density of the Fluid = Specific Gravity of the fluid x Density of water = 1.25 x 62.4
Density of the Fluid p = 78 lbm/ft^3
Difference in pressure as we got the differential height, dP = p x g x h dP = (78 lbm/ft^3) x (32.174 ft/s^2) x (28/12 ft) [ 1 lbf / 32.174 ft/s^2] [1 ft^2 /
144in^2]
Difference in pressure = 1.26 psia
(a) Pressure in the arm that is at Higher
P = Atmospheric Pressure - Pressure difference = 12.7 - 1.26 = 11.44 psia
(b) Pressure in the tank that is at Lower
P = Atmospheric Pressure + Pressure difference = 12.7 + 1.26 = 13.96psia
Initial speed = 2√10 m/s
<h3>Further explanation </h3>
Linear motion consists of 2: constant velocity motion with constant velocity and uniformly accelerated motion with constant acceleration
An equation of uniformly accelerated motion
V = vo + at
Vt² = vo² + 2a (x-xo)
x = distance on t
vo / vi = initial speed
vt / vf = speed on t / final speed
a = acceleration
vf=20 m/s
d = 60 m
a = 3 m/s²
Answer:
Sun
Explanation:
The sun is Bigger than everything else.
Answer:
Current through each phase Vp = 2.2A
Total three phase power Pt= 1.45kW
Power factor of the load pf = 1
Explanation:
i) Find current through each phase
Vp =220V (rms)
Z =100 Ω
I = Vp/Z
= 220/100
= 2.2A
ii) Find the total three phase power
for a resistive load, Power, P = VI
Power for each phase is given as:
P = 220 * 2.2
= 484 W
Total power TP =3* P
=484*3
= 1452W
=1.45kW
iii) Find the power factor of the load
Phase angle for a resistive load is 0.
α= 0
Hence, power factor of load = cos α
pf = cos 0
pf = 1
Answer:
a)η = 69.18 %
b)W= 1210 J
c)P=3967.21 W
Explanation:
Given that
Q₁ = 1749 J
Q₂ = 539 J
From first law of thermodynamics
Q₁ = Q₂ +W
W=Work out put
Q₂=Heat rejected to the cold reservoir
Q₁ =heat absorb by hot reservoir
W= Q₁- Q₂
W= 1210 J
The efficiency given as
η = 69.18 %
We know that rate of work done is known as power
P=3967.21 W
