Answer:
(a) BP = 11.99 KPa
(b) h = 2 m
Explanation:
(a)
Since, the fluid pressure and blood pressure balance each other. Therefore:
BP = ρgh
where,
BP = Blood Pressure
ρ = density of fluid = 1020 kg/m³
g = acceleration due to gravity = 9.8 m/s²
h = height of fluid = 1.2 m
Therefore,
BP = (1020 kg/m³)(9.8 m/s²)(1.2 m)
<u>BP = 11995.2 Pa = 11.99 KPa</u>
(b)
Again using the equation:
P = ρgh
with data:
P = Gauge Pressure = 20 KPa = 20000 Pa
ρ = density of fluid = 1020 kg/m³
g = acceleration due to gravity = 9.8 m/s²
h = height of fluid = ?
Therefore,
20000 Pa = (1020 kg/m³)(9.8 m/s²)h
<u>h = 2 m</u>
Answer:
1200KJ
Explanation:
The heat dissipated in the rotor while coming down from its running speed to zero, is equal to three times its running kinetic energy.
P (rotor-loss) = 3 x K.E
P = 3 x 300 = 900 KJ
After coming to zero, the motor again goes back to running speed of 1175 rpm but in opposite direction. The KE in this case would be;
KE = 300 KJ
Since it is in opposite direction, it will also add up to rotor loss
P ( rotor loss ) = 900 + 300 = 1200 KJ
Answer:
square cross section. The bar is made of a 7075-T6 aluminum alloy which has a yield strength of 500 MPa, a tensile strength of 575 MPa, and a fracture toughness of 27.5 MPaâm.
Required:
a. What is the nominal maximum tensile stress on the bar?
b. If there were an initial 1.2 mm deep surface crack on the right surface of the bar, what would the critical stress needed to cause instantaneous fast fracture of the bar be?
Answer:
a) 
b)
Explanation:
Given that:
diameter d = 12 in
thickness t = 0.25 in
the radius = d/2 = 12 / 2 = 6 in
r/t = 6/0.25 = 24
24 > 10
Using the thin wall cylinder formula;
The valve A is opened and the flowing water has a pressure P of 200 psi.
So;
b)The valve A is closed and the water pressure P is 250 psi.
where P = 250 psi
The free flow body diagram showing the state of stress on a volume element located on the wall at point B is attached in the diagram below
The impact behavior of plastic materials is strongly dependent upon the temperature. At high temperatures, materials are more ductile and have high impact toughness. At low temperatures, some plastics that would be ductile at room temperature become brittle.
