Answer:
(a) The unit of 70.5 is lbm/ft^3 and the unit of 8.27×10^-7 is in^2/lbf
(b) density = 0.1206g/cm^3
(c) rho = 0.1206exp(8.27×10^-7P)
Step-by-step explanation:
(a) The unit of 70.5 is the same as the unit of rho which is lbm/ft^3. The unit of 8.27×10^-7 is the inverse of the unit of P (lbf/in^2) because exp is found of a constant. Therefore, the unit of 8.27×10^-7 is in^2/lbf
(b) P = 9×10^6N/m^2
rho = 70.5exp(8.27×10^-7× 9×10^6) = 70.5exp7.443 = 70.5×1.71 = 120.6kg/m^3
rho = 120.6kg/m^3 × 1000g/1kg × 1m^3/10^6cm^3 = 0.1206g/cm^3
(c) Formula for rho (g/cm^3) as a function of P (N/m^2) is
rho = 0.1206exp(8.27×10^-7P) (the unit of 0.1206 is g/cm^3)
Answer:solo quiero puntos :(
Step-by-step explanation:
Step-by-step explanation:
(X1,Y1) = (-2,2)
(X2,Y2) = (0,1)
• Find slope.
m = (Y2 - Y1)/(X2 - X1)
m = (1 - 2)/(0 - (-2))
m = -1/2
The answer is C.
3.50x + 5 < 30
Ellie wants to spend less than 30, so it’ll be less than.
Answer:
A(t) = 300 -260e^(-t/50)
Step-by-step explanation:
The rate of change of A(t) is ...
A'(t) = 6 -6/300·A(t)
Rewriting, we have ...
A'(t) +(1/50)A(t) = 6
This has solution ...
A(t) = p + qe^-(t/50)
We need to find the values of p and q. Using the differential equation, we ahve ...
A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50
0 = 6 -p/50
p = 300
From the initial condition, ...
A(0) = 300 +q = 40
q = -260
So, the complete solution is ...
A(t) = 300 -260e^(-t/50)
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The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.
