For the number less than 0.1 such as 0.006, the zeroes to the right of the decimal point but before the first non zero digit show the decimal place of the first significant digit.
- The number that is given as digits is established using significant figures.
- Any two non-zero digits that are separated by a zero are significant figure.
- Every zero that is both to the right and left of a non-zero digit and the decimal point is not significant figure.
- The quantity of significant figures frequently reveals the degree of measurement accuracy. From the first non-zero digits in the figure, we may determine the number of significant figures.
There is only one significant figure in the provided number 0.06. The decimal place of the first digit is indicated by the zeros that appear to the right of the decimal point but before the first non-zero digit.
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Answer:
x(t) = −39e
−0.03t + 40.
Explanation:
Let V (t) be the volume of solution (water and
nitric acid) measured in liters after t minutes. Let x(t) be the volume of nitric acid
measured in liters after t minutes, and let c(t) be the concentration (by volume) of
nitric acid in solution after t minutes.
The volume of solution V (t) doesn’t change over time since the inflow and outflow
of solution is equal. Thus V = 200 L. The concentration of nitric acid c(t) is
c(t) = x(t)
V (t)
=
x(t)
200
.
We model this problem as
dx
dt = I(t) − O(t),
where I(t) is the input rate of nitric acid and O(t) is the output rate of nitric acid,
both measured in liters of nitric acid per minute. The input rate is
I(t) = 6 Lsol.
1 min
·
20 Lnit.
100 Lsol.
=
120 Lnit.
100 min
= 1.2 Lnit./min.
The output rate is
O(t) = (6 Lsol./min)c(t) = 6 Lsol.
1 min
·
x(t) Lnit.
200 Lsol.
=
3x(t) Lnit.
100 min
= 0.03 x(t) Lnit./min.
The equation is then
dx
dt = 1.2 − 0.03x,
or
dx
dt + 0.03x = 1.2, (1)
which is a linear equation. The initial condition condition is found in the following
way:
c(0) = 0.5% = 5 Lnit.
1000 Lsol.
=
x(0) Lnit.
200 Lsol.
.
Thus x(0) = 1.
In Eq. (1) we let P(t) = 0.03 and Q(t) = 1.2. The integrating factor for Eq. (1) is
µ(t) = exp Z
P(t) dt
= exp
0.03 Z
dt
= e
0.03t
.
The solution is
x(t) = 1
µ(t)
Z
µ(t)Q(t) dt + C
= Ce−0.03t + 1.2e
−0.03t
Z
e
0.03t
dt
= Ce−0.03t +
1.2
0.03
e
−0.03t
e
0.03t
= Ce−0.03t +
1.2
0.03
= Ce−0.03t + 40.
The constant is found using x(t) = 1:
x(0) = Ce−0.03(0) + 40 = C + 40 = 1.
Thus C = −39, and the solution is
x(t) = −39e
−0.03t + 40.
Answer: they will meet and make a biger wave then seperat
Explanation: When two or more waves meet, they interact with each other. The interaction of waves with other waves is called wave interference. Wave interference may occur when two waves that are traveling in opposite directions meet. The two waves pass through each other, and this affects their amplitude.
Something that is special to you or event that means alot to you
1) The question contains an unknown unit
The number 8908.8 has to be in units of mass: for example, kg or grams.
Here you indicated L.
I am going to work assuming that L is a mass unit. So you can see the way to solve the problem, but you have to verifiy the real unit of the statement and substitute with it.
With that in mind you can find the density of the liquid from:
density = mass / volume
2) Calculate the volume.
The volume of the liquid is the volume of the vessel, because it is filled.
The volume of the vessel is calculated from the formula of volume for a rectantular prism.
Volume of a rectangular prism = area of the base * height = side * side * height
=> Volume = 10 dm * 4 dm * 6 dm = 240 dm^3 = 240 liter
3) Calculate the density:
density = mass /volume = 8,908.8 L / 240 liter = 37.12 L / dm^3
Answer: 37.12 L / dm^3
