Answer:
V=1601gal
Explanation:
Hello! This problem is solved as follows,
First we must raise the equation that defines the pressure at the bottom of the tank with the purpose of finding the height that olive oil reaches.
This is given as the sum due to the atmospheric pressure (1atm = 101.325kPa), and the pressure due to the weight of the olive oil, taking into account the above, the following equation is inferred.
P=Poil+Patm
P=total pressure or absolute pressure=26psi=179213.28Pa
Patm= the atmospheric pressure =101325Pa
Poil=pressure due to the weight of olive oil=0.86αgh
α=density of water=1000kg/m^3
g=gravity=9.81m/s^2
h= height that olive oil reaches
solving
P=Poil+Patm
P=Patm+0.86αgh
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Now we can use the equation that defines the volume of a cylinder.
V=
D=3ft=0.9144m
h=9.23m
solving
finally we use conversion factors to find the volume in gallons
Answer:
0.25 J/K
Explanation:
Given data in given question
heat (Q) = 100 J
temperature (T) = 400 K
to find out
the change in entropy of the given system
Solution
we use the entropy change equation here i.e
ΔS = ΔQ / T ...................a
Now we put the value of heat (Q) and Temperature (T) in equation a
ΔS is the entropy change, Q is heat and T is the temperature,
so that
ΔS = 100/400 J/K
ΔS = 0.25 J/K
Answer:
my_mul:
.globl my_mul
my_mul:
//Multiply X0 and X1
// Does not handle negative X1!
// Note : This is an in efficient way to multipy!
SUB SP, SP, 16 //make room for X19 on the stack
STUR X19, [SP, 0] //push X19
ADD X19, X1, XZR //set X19 equal to X1
ADD X9 , XZR , XZR //set X9 to 0
mult_loop:
CBZ X19, mult_eol
ADD X9, X9, X0
SUB X19, X19, 1
B mult_loop
mult_eol:
LDUR X19, [SP, 0]
ADD X0, X9, XZR // Move X9 to X0 to return
ADD SP, SP, 16 // reset the stack
BR X30
Explanation:
