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vodomira [7]
3 years ago
8

In 1985, there were 285 cell phone subscribers in the small town of Glenwood. The number of subscribers increased by 75% per yea

r after 1985. How many cell phone subscribers were in Glenwood in 1990? Years 1986 (x = 1) 1987 (x = 2) 1988 (x = 3) Number of cell phone users 498 872 1527 a. 1800 c. 2672 b. 4677 d. 8186 Please select the best answer from the choices providednd A
Physics
2 answers:
Amiraneli [1.4K]3 years ago
4 0

Answer:

b. (4677)

Explanation:

You can apply the same equation we use to calculate the compound interest but with phones.

Look:

M = C.(1+i)^n

M = the total amount after 5 years

C = the firts amout of phones (285)

i = how much this number gets bigger (75%)

n = the number of periods (5 years)

So...

M = 285 . (1 + 0,75)^5

M = 285 . 1,75^5

M = 285 . 16.4

M = 4677

DedPeter [7]3 years ago
4 0

Answer:

b

Explanation:

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8 0
4 years ago
A mass of 0.75 kilograms is attached to a spring/mass oscillator. A force of 5 newtons is required to stretch the spring 0.5 met
zlopas [31]

Answer:

b > 66.41 kg/s

Explanation:

The spring force F = -kx, where k = spring constant, the damping force f = -bv. The net force F' = F + f

F + f = ma

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Re-arranging the equation, we have

So, md²x/dt² + bdx/dt + kx = 0

Dividing through by m, we have

d²x/dt² + (b/m)dx/dt + (k/m)x = 0

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D² + (b/m)D + (k/m) = 0

Using the quadratic formula, we find D.

D = \frac{-(b/m) +/- \sqrt{(b/m)^{2} - 4k/m} }{2}

For an overdamped system,

(b/m)^{2} - 4k/m} >   0

(b/m)^{2} >   4k/m}\\(b/m) >   \sqrt{4k/m}} \\(b/m) >   2\sqrt{k/m}} \\b >   2\sqrt{km}}

Now, k = F/x. Since the weight of the object causes the spring to stretch a distance of 0.5 m, k = mg/x where m = mass of object = 0.75 kg, g = 9.8 m/s² and x = x₀ =0.5 m.

Substituting k = mg/x into the inequality for b, we have

b > 2√{(mg/x₀)m}

b > 2√{(m²g/x₀)}

b > 2m√{g/x₀)}

b > 2 × 0.75 kg√{9.8 m/s²/0.5 m)}

b > 1.5 kg√{19.6/s²)}

b > 1.5 kg × 4.427/s

b > 66.41 kg/s

6 0
3 years ago
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