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ikadub [295]
3 years ago
13

A radio station broadcasts a signal over an area with a 120-mile diameter. The population density of the area is 100 people per

square mile.
How many people are within receiving distance of the broadcast signal?

Use 3.14 for pi.

Select from the drop-down menu to correctly complete the statement.
There are about____ people within broadcast distance of the radio station.
Mathematics
2 answers:
elena-s [515]3 years ago
8 0

I did the test, and i got the answer of 1, 130, 400.

Hope this helps!!!!! ;)

Rama09 [41]3 years ago
3 0
R = 120 / 2 = 60 miles
Density of the area:  100 people per sq. mile
A = r² π = 60² · 3.14 = 3600 · 3.14 = 11,304 sq. miles
11,304 · 100 = 1,030,400 
Answer:
There are about 1,030,400 people within broadcast distance of the radio station.
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A 15 kilogram object is suspended from the end of a vertically hanging spring stretches the spring 1/3 meters. At time t = 0, th
Yuri [45]

Answer:

15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = ± 170 cos(5t)

y(0)=0, y'(0)=0

Step-by-step explanation:

See the attached image

This problem involves Newton's 2nd Law which is: ∑F = ma, we have that the acting forces on the mass-spring system are: F_{r} (t) that correspond to the force of resistance on the mass by the action of the spring and F(t) that is an external force with unknown direction (that does not specify in the enounce).

For determinate F_{r} (t) we can use Hooke's Law given by the formula F_{r} (t) = k y(t) where k correspond to the elastic constant of the spring and y(t) correspond to  the relative displacement of the mass-spring system with respect of his rest state.

We know from the problem that an 15 Kg mass stretches the spring 1/3 m so we apply Hooke's law and obtain that...

k = \frac{F_{r}}{y} = \frac{mg}{y} = \frac{15 Kg (9.81 \frac{m}{s^{2} } )}{\frac{1}{3} m}  = 441.45 \frac{N}{m}

Now we apply Newton's 2nd Law and obtaint that...

F_{r} (t) ± F(t) = ma(t)

F_{r} (t) = ky(t) = 441.45y(t)

F(t) = 170 cos(5t)

m = 15 kg

a(t) = \frac{d^{2}y(t)}{dt^{2} }

Finally... 15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = ± 170 cos(5t)

We know from the problem that there's not initial displacement and initial velocity, so... y(0)=0 and y'(0)=0

Finally the Initial Value Problem that models the situation describe by the problem is

\left \{ 15\frac{d^{2}y(t)}{dt^{2} }  - 441.45y(t) = \frac{+}{} 170 cos(5t) \atop {y(0)=0, y'(0)=0\right.

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3 years ago
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Answer

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2.(-6,0) (C)

Step-by-step explanation:

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