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Vlada [557]
3 years ago
12

Find a function that models the simple harmonic motion having the given properties. Assume that the displacement is zero at time

t = 0.
amplitude 16 cm, period 3 s



Would the answer bey = 16sin(\frac{2\pi }{3}t)
Mathematics
1 answer:
Svetlanka [38]3 years ago
5 0

Answer:

y = 16sin(2π/ 3)t.

Step-by-step explanation:

When displacement = 0 at time = 0 we use the sine function.

The formula is y = A sin(2πft)  where A = the amplitude, f = frequency which  is 1/period and t = the time.

So for this SHM the function is

y = 16sin(2π/ 3) t.

You are correct.

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The annual growth rate of energy utilization in the world was 3.5% per year in the period between 1950 and 1973. How long would
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Step-by-step explanation:

4 0
3 years ago
For each rectangle below, write a linear equation that represents the area y of the rectangle. Solve this system of two linear e
Hatshy [7]

(5,24)

  • The two areas are the same.

  • To find the area, we multiply the side lengths. Y=area

Rectangle 1: side lengths 4 and (x+1)

y=4(x+1)= 4x+4

Rectangle 2: side lengths 3 and (2x-2)

y=3(2x-2)= 6x-6

  • Since the two areas are same, we can conclude that

4x+4=6x-6

  • Now, simplify.

-2x=-10, x=5

  • Since x is 5, we can plug it into the equations to find y.

Option 1 with rectangle 1: y=4(5)+4, y=24

Option 2 with rectangle 2: y=6(5)-5, y=24

I graphed the linear equation on desmos.

5 0
3 years ago
The radius of a circle is changing at the rate of 1/π inches per second. At what rate, in square inches per second, is the circl
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Answer:

10 square inches per second.

Step-by-step explanation:

The radius of the circle is given by the equation:

r(t) = (1/π  in/s)*t

Where time in seconds.

Remember that the area of a circle of radius R is written as:

A = π*R^2

Then the area of our circle will be:

A(t) = π*( (1/π  in/s)*t)^2 = π*(1/π  in/s)^2*(t)^2

Now we want to find the rate of change (the first derivation of the area) when the radius is equal to 5 inches.

Then the first thing we need to do is find the value of t such that the radius is equal to 5 inches.

r(t) = 5 in =  (1/  in/s)*t

       5in*(π s/in) = t

        5*π s = t

So the radius will be equal to 5 inches after 5*π seconds, let's remember that.

Now let's find the first derivate of A(t)

dA(t)/dt = A'(t) = 2*(π*(1/π  in/s)^2*t = (2*π*t)*(1/π  in/s)^2

Now we need to evaluate this in the time such that the radius is equal to 5 inches, we will get:

A'(5*π s) = (2*π*5*π s)*((1/π  in/s)^2

              = (10*π^2  s)*(1/π^2  in^2/s^2) = 10 in^2/s

The rate of change is 10 square inches per second.

4 0
3 years ago
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