Answer:
a) 1253 kJ
b) 714 kJ
c) 946 C
Explanation:
The thermal efficiency is given by this equation
η = L/Q1
Where
η: thermal efficiency
L: useful work
Q1: heat taken from the heat source
Rearranging:
Q1 = L/η
Replacing
Q1 = 539 / 0.43 = 1253 kJ
The first law of thermodynamics states that:
Q = L + ΔU
For a machine working in cycles ΔU is zero between homologous parts of the cycle.
Also we must remember that we count heat entering the system as positiv and heat leaving as negative.
We split the heat on the part that enters and the part that leaves.
Q1 + Q2 = L + 0
Q2 = L - Q1
Q2 = 539 - 1253 = -714 kJ
TO calculate a temperature for the heat sink we must consider this cycle as a Carnot cycle. Then we can use the thermal efficiency equation for the Carnot cycle, this one uses temperatures:
η = 1 - T2/T1
T2/T1 = 1 - η
T2 = (1 - η) * T1
The temperatures must be given in absolute scale (1453 C = 1180 K)
T2 = (1 - 0.43) * 1180 = 673 K
673 K = 946 C
Answer:
Qcd=0.01507rad
QT= 0.10509rad
Explanation:
The full details of the procedure and answer is attached.
Answer:
investment 10 years from now is $1,238,000
.
Explanation:
given data
sum = $500,000
rate = 12% =0.12
total time = 10 year
solution
as present value After 2 years from now is $500,000
so time period is now = 8 year ( 10 - 2 )
so we apply future value formula that is
Future value = present value ×
............1
put here value we get
Future value = $500,000 ×
Future value = $500,000 × 2.476
Future value = $1,238,000
so investment 10 years from now is $1,238,000
.
Answer:
a)R= sqrt( wt³/12wt)
b)R=sqrt(tw³/12wt)
c)R= sqrt ( wt³/12xcos45xwt)
Explanation:
Thickness = t
Width = w
Length od diagonal =sqrt (t² +w²)
Area of raectangle = A= tW
Radius of gyration= r= sqrt( I/A)
a)
Moment of inertia in the direction of thickness I = w t³/12
R= sqrt( wt³/12wt)
b)
Moment of inertia in the direction of width I = t w³/12
R=sqrt(tw³/12wt)
c)
Moment of inertia in the direction of diagonal I= (w t³/12)cos 45=( wt³/12)x 1/sqrt (2)
R= sqrt ( wt³/12xcos45xwt)