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3 years ago

A closed cylindrical tank had a diameter and height 7m what is the capacity of the tank

Mathematics

1 answer:

bezimeni [28]3 years ago

3 0

Area of a cylinder is pi*r^2*h

The radius is half the diameter.

So the radius is 3.5

3.14*3.5^2*7=269.255 cubic meters.

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A mass weighing 16 pounds stretches a spring (8/3) feet. The mass is initially released from rest from a point 2 feet below the

mezya [45]

Answer with Step-by-step explanation:

Let a mass weighing 16 pounds stretches a spring $\frac{8}{3}$ feet.

Mass= $m=\frac{W}{g}$

Mass= $m=\frac{16}{32}$

$g=32 ft/s^2$

Mass,m= $\frac{1}{2}$ Slug

By hook's law

$w=kx$

$16=\frac{8}{3} k$

$k=\frac{16\times 3}{8}=6 lb/ft$

$f(t)=10cos(3t)$

A damping force is numerically equal to 1/2 the instantaneous velocity

$\beta=\frac{1}{2}$

Equation of motion :

$m\frac{d^2x}{dt^2}=-kx-\beta \frac{dx}{dt}+f(t)$

Using this equation

$\frac{1}{2}\frac{d^2x}{dt^2}=-6x-\frac{1}{2}\frac{dx}{dt}+10cos(3t)$

$\frac{1}{2}\frac{d^2x}{dt^2}+\frac{1}{2}\frac{dx}{dt}+6x=10cos(3t)$

$\frac{d^2x}{dt^2}+\frac{dx}{dt}+12x=20cos(3t)$

Auxillary equation

$m^2+m+12=0$

$m=\frac{-1\pm\sqrt{1-4(1)(12)}}{2}$

$m=\frac{-1\pmi\sqrt{47}}{2}$

$m_1=\frac{-1+i\sqrt{47}}{2}$

$m_2=\frac{-1-i\sqrt{47}}{2}$

Complementary function

$e^{\frac{-t}{2}}(c_1cos\frac{\sqrt{47}}{2}+c_2sin\frac{\sqrt{47}}{2})$

To find the particular solution using undetermined coefficient method

$x_p(t)=Acos(3t)+Bsin(3t)$

$x'_p(t)=-3Asin(3t)+3Bcos(3t)$

$x''_p(t)=-9Acos(3t)-9sin(3t)$

This solution satisfied the equation therefore, substitute the values in the differential equation

$-9Acos(3t)-9Bsin(3t)-3Asin(3t)+3Bcos(3t)+12(Acos(3t)+Bsin(3t))=20cos(3t)$

$(3B+3A)cos(3t)+(3B-3A)sin(3t)=20cso(3t)$

Comparing on both sides

$3B+3A=20$

$3B-3A=0$

Adding both equation then, we get

$6B=20$

$B=\frac{20}{6}=\frac{10}{3}$

Substitute the value of B in any equation

$3A+10=20$

$3A=20-10=10$

$A=\frac{10}{3}$

Particular solution, $x_p(t)=\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)$

Now, the general solution

$x(t)=e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)$

From initial condition

x(0)=2 ft

x'(0)=0

Substitute the values t=0 and x(0)=2

$2=c_1+\frac{10}{3}$

$2-\frac{10}{3}=c_1$

$c_1=\frac{-4}{3}$

$x'(t)=-\frac{1}{2}e^{-\frac{t}{2}}(c_1cos(\frac{\sqrt{47}t}{2})+c_2sin(\frac{\sqrt{47}t}{2})+e^{-\frac{t}{2}}(-c_1\frac{\sqrt{47}}{2}sin(\frac{\sqrt{47}t}{2})+\frac{\sqrt{47}}{2}c_2cos(\frac{\sqrt{47}t}{2})-10sin(3t)+10cos(3t)$

Substitute x'(0)=0

$0=-\frac{1}{2}\times c_1+10+\frac{\sqrt{47}}{2}c_2$

$\frac{\sqrt{47}}{2}c_2-\frac{1}{2}\times \frac{-4}{3}+10=0$

$\frac{\sqrt{47}}{2}c_2=-\frac{2}{3}-10=-\frac{32}{3}$

$c_2==-\frac{64}{3\sqrt{47}}$

Substitute the values then we get

$x(t)=e^{-\frac{t}{2}}(-\frac{4}{3}cos(\frac{\sqrt{47}t}{2})-\frac{64}{3\sqrt{47}}sin(\frac{\sqrt{47}t}{2})+\frac{10}{3}cos(3t)+\frac{10}{3}sin(3t)$

8 0

3 years ago

Jim deposits 6000 into an account that pays simple interest at a rate of 2% per year. How much interest will he be paid in the f

Alex73 [517]

Answer:

$600 interest will be paid in the first 5 years

Step-by-step explanation:

Simple interest is calculated only on the principal or initial amount deposited. Jim deposits 6000 into an account that pays simple interest at a rate of 2% per year. The formula for simple interest is expressed as

I = PRT/100

Where

I represents interest after the given time.

R represents the interest loan on the principal.

T represent number of time in years.

P represents principal or initial amount deposited.

From the information given,

R = 2

P = 6000

T = 5

I = (6000 × 2 × 5)/100 = $600

6 0

3 years ago

Isaac invested $77,000 in an account paying an interest rate of 4.6% compounded continuously. Assuming no deposits or withdrawal

zlopas [31]

Answer:

132,090.30

Step-by-step explanation:

r = R/100

r = 4.6%/100

r = 0.046 per year,

A = P(1 + r/n)nt

A = 77,000.00(1 + 0.046/1)(1)(12)

A = $ 132,090.30

3 0

2 years ago

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Show the steps to solve the system of equation. x-2y+3z=-4 , 3x+y-z=0 , 2x+3y-5z=1

Taya2010 [7]

Answer:

x= -1, y = 6, z = 3

Step-by-step explanation:

let me solve it with Gaussian Elimination

8 0

3 years ago

7/5 x -3/12 + 7/6 + 7/5 x 1/4

Studentka2010 [4]

Answer:

7/6 or 1.166666667

Step-by-step explanation:

I solved it on calculator

To 1 d.p 1.2

6 0

3 years ago

Read 2 more answers