Answer:
0.4 gallons per second
Explanation:
A function shows the relationship between an independent variable and a dependent variable.
The independent variable (x values) are input variables i.e. they don't depend on other variables while the dependent variable (y values) are output variables i.e. they depend on other variables.
The rate of change or slope or constant of proportionality is the ratio of the dependent variable (y value) to the independent variable (x value).
Given that the garden hose fills a 2-gallon bucket in 5 seconds. The dependent variable = g = number of gallons, the independent variable = t = number of seconds.
Constant of proportionality = g / t = 2 / 5 = 0.4 gallons per second
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
Answer:
W = 1 mJ
V_new = 2000 V
W_new = 2 mJ
The extra energy came from the work done from moving the plates
Explanation:
We are given;
Capacitance; C = 1000pF = 10^(-9) F
Voltage; V = 1000V
Now,formula for stored energy in a parallel plate capacitor is given by;
W = (1/2)CV² = (1/2)(10^(-9))(1000²) = 0.5 mJ
However, in this case, it's W = 0.5 x 2 = 1 mJ since parallel-plate capacitor with air dielectric
When plates are moved and distance between plates is doubled(net charge is the same), thus we can calculate voltage from;
Q = CV
Since, C_new = C/2
Thus,
Q = (C/2)V_new
V_new = 2Q/C
Thus, V_new = 2V
Thus, V_new = 2 x 1000 = 2000 V
Now,
W_new = (1/2)C_new•(V_new)² = (1/2) (0.5C)•(2V)² = CV² = 2W = 2 x 1 = 2mJ
Answer:
a) -505.229 kJ/Kg
b) -1.724 kJ/kg
Explanation:
T1 = 400°C
P1 = 3 MPa
P2 = 125 kPa
work output = 530 kJ/kg
surrounding temperature = 20°C = 293 k
<u>A) Calculate heat transfer from Turbine to surroundings </u>
Q = h2 + w - h1
h ( enthalpy )
h1 = 3231.229 kj/kg
enthalpy at P2
h2 = hg = 2676 kj/kg
back to equation 1
Q = 2676 + 50 - 3231.229 = -505.229 kJ/Kg ( i.e. heat is lost )
<u>b) Entropy generation </u>
entropy generation = Δs ( surrounding ) + Δs(system)
= - 505.229 / 293 + 0
= -1.724 kJ/kg
