Answer:
Logic NOR Gate Equivalent
The Boolean expression for a logic NOR gate is denoted by a plus sign, ( + ) with a line or Overline, ( ‾‾ ) over the expression to signify the NOT or logical negation of the NOR gate giving us the Boolean expression of: A+B = Q.
Answer:
88.18 W
Explanation:
The weight of the boy is given as 108 lb
Change to kg =108*0.453592= 48.988 kg = 49 kg
The slope is given as 6% , change it to degrees as
6/100 =0.06
tan⁻(0.06)= 3.43°
The boy is travelling at a constant speed up the slope = 7mi/hr
Change 7 mi/h to m/s
7*0.44704 =3.13 m/s
Formula for power P=F*v where
P=power output
F=force
v=velocity
Finding force
F=m*g*sin 3.43°
F=49*9.81*sin 3.43° =28.17
Finding the power out
P=28.17*3.13 =88.18 W
Answer:
1. High friction
2. High extrusion temperature
Explanation:
Surface cracking on extruded products are defects or breakage on the surface of the extruded parts. Such cracks are inter granular.
Surface cracking defects arises from very high work piece temperature that develops cracks on the surface of the work piece. Surface cracking appears when the extrusion speed is very high, that results in high strain rates and generates heat.
Other factors include very high friction that contributes to surface cracking an d chilling of the surface of high temperature billets.
Answer:
Explanation:
First, we will find actual properties at given inlet and outlet states by the use of steam tables:
AT INLET:
At 4MPa and 350°C, from the superheated table:
h₁ = 3093.3 KJ/kg
s₁ = 6.5843 KJ/kg.K
AT OUTLET:
At P₂ = 125 KPa and steam is saturated in vapor state:
h₂ = 
= 2684.9 KJ/kg
Now, for the isentropic enthalpy, we have:
P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K
Since s₂ is less than 
and greater than 
at 125 KPa. Therefore, the steam is in a saturated mixture state. So:
Now, we will find 
(enthalpy at the outlet for the isentropic process):
Now, the isentropic efficiency of the turbine can be given as follows:
