Answer:
I want to believe the program is to be written in java and i hope your question is complete. The code is in the explanation section below
Explanation:
import java.util.Date;
public interface Downloadable {
//abstract methods
public String getUrl();
public Date getLastDownloadDate();
}
Answer:
(a) 1/L∫Vdt; integral t [0,1]
(b) 1/L∫Vdt; integral t [ 1, infinity]
Explanation:
An Inductor current I, flowing through an inductor depends on the voltage, V, across the inductor and the inductance, L, of the inductor. The switch 1, 2 timing varies the voltage V with time t
The expression for inductor current is given as:
I= 1/L∫Vdt,
where I is equal to the current flowing through the inductor, L is equal to the inductance of the inductor, and V is equal to the voltage across the inductor.
The formula can also be written as:
I= I0 + 1/L∫Vdt, where I is inductor current at time t, and io is inductor current at t = 0. Time can be varied by controlling the switch
Answer:
Each of the metal specimens HAS an indentation near the center to ensure that the fracture point would occur in this region. Tension tests WERE CONDUCTED as follows. Two pieces of reflective tape WERE PLACED approximately 1 inch apart in the center of the specimen where the indentation 4 WAS LOCATED. The width and the thickness of the specimen at this location WAS MEASURED using a Vernier caliper. Then the specimen WAS SECURED in the MTS Load Frame. A laser extensometer WAS PLACED into position to measure the deformation of the specimen. The laser extensometer WAS USED to measure the original distance between the pieces of reflective tape. The MTS WAS SET to elongate the specimen one tenth of an inch every minute.
Choose a quality one, and don't use it as necessary
Answer:
A) 3783.952 kJ/kg
B) 34.5%
C) 2476.67 kJ/kg
Explanation:
A) Determine the rate of heat addition entering the first-stage turbine
Qin ( 1st stage ) = ( h1 - h6 ) + ( h3 - h2 )
= ( 3321.4 - 164.07 ) + ( 3438.566 - 2811.944 )
= 3783.952 kJ/kg
<em>h values are gotten from super heated table </em>
B) Determine the thermal efficiency
n = 34.5%
attached below
C) Rate of heat transfer from working fluid passing through the condenser to the cooling water
Qout = h4 - h5
= 2628.2 - 151.53
= 2476.67 kJ/kg