Answer: 
= 28800 Pa or 28.8 kPa
Explanation: To determine the pressure of a liquid in a rotating tank,it is used:
p = 
- γfluid . z + c
where:
is the liquid's density
w is the angular velocity
r is the radius
γfluid.z is the pressure variation due to centrifugal force.
For this question, the difference between a point on the circumference and a point on the axis will be:
= 
- γfluid.
- (
- γfluid.
)
= 
- γfluid( -)
Since there is no variation in the z-axis, z = 0 and that the density of oil is 0.9.10³kg/m³:
=
= 28800
The difference in pressure between two points, one on the circumference and the other on the axis is = 28800 Pa or 28.8 kPa
Answer:
248.756 mV
49.7265 µA
Explanation:
The Thevenin equivalent source at one terminal of the bridge is ...
voltage: (100 V)(1000/(1000 +1000) = 50 V
impedance: 1000 || 1000 = (1000)(1000)/(1000 +1000) = 500 Ω
The Thevenin equivalent source at the other terminal of the bridge is ...
voltage = (100 V)(1010/(1000 +1010) = 100(101/201) ≈ 50 50/201 V
impedance: 1000 || 1010 = (1000)(1010)/(1000 +1010) = 502 98/201 Ω
__
The open-circuit voltage is the difference between these terminal voltages:
(50 50/201) -(50) = 50/201 V ≈ 0.248756 V . . . . open-circuit voltage
__
The current that would flow is given by the open-circuit voltage divided by the sum of the source resistance and the load resistance:
(50/201 V)/(500 +502 98/201 +4000) = 1/20110 A ≈ 49.7265 µA
I believe it’s True, but please correct me if I’m wrong!
Answer:
See the attached file for the answer.
Explanation:
See the attached file for the explanation
Answer:
<u>the thickness required would be 12 inch HMA and granular base layer of 6 inches</u>
Explanation:
structural number = 4.5
stone base course material coefficient = 0.13
hma material layer coefficient = 0.40
drainage coefficient = 0.90
we will use layered analysis procedure to get thickness
D1 >= sN1/a1
when we cross multiply,
sN1 = a1D1 >=sN1
D2 >= -sN2-sN1/a2m2
sN2* + sN1* >= sN2
D3 >= sN3-(sN1*+sN2*)/a2m2
where a1,a2,a3 = layer coefficient
d1 d2 d3 = actual thickness
m2,m3 = coefficient of base
a1 = 0.4
a1 = 0.13
sN = 4.5
m2 = 0.9
D1 >= sN1/a1 = 4.5/0.4
= 11.25
thickness of surface = 12 inches
a1D1 = 0.4x12 = 4.8
we have value of sN2 = 5.5
(5.5 -4.8)/(0.13*0.9)
= 0.7/0.117
= 5.9829 inches
approximately 6 inches
so the pavement will have 12inch HMA surface and 6 inches granular base layer.
