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3 years ago

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A city’s bus line is used more as the urban population density increases. The more people in an area, the more likely bus lines

will be utilized. The population of City A, in Texas, is 29,455 and covers an area of 19.42 mi2. The population of City B, in Alabama, is 34,890 and covers an area of 25.4 mi2. Which city’s residents are more likely to use public buses? Explain

2 answers:

Nonamiya [84]3 years ago

8 0

Answer:

City bus A

Explanation:

I just took the test

Aleksandr-060686 [28]3 years ago

7 0

Answer:

City A residents

Explanation:

First we need to calculate the population density of each city. This is calculated by dividing the total population by the total area of the city.

City A has 29455 people and 19.42mi2 of area, so the density is 1516.7 people per mile squared.

City B has 34890 people and 25.4 mi2 of area, so the density is 1373.6 people per mile squared.

If the city with bigger people density is more likely to use bus lines, the city A is the one whose people are more likely to use bus lines, as it has a bigger people density.

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A disk-shaped merry-go-round of radius 2.63 m and mass 152 kg rotates freely with an angular speed of 0.526 rev/s. A 51.7 kg per

zalisa [80]

Explanation:

Given

radius $r=2.63 m$

$N=0.526 rev/s$

$\omega =3.30 rad/s$

mass disc $M=152 kg$

mass of person $m=51.7 kg$

velocity of Person $v=2.76 m/s$

moment of inertia $I=Mr^2$

$I=0.5\times 152\times 2.63^2=827.64 kg-m^2$

Initial angular momentum

$L_i=I\omega +mvr$

$L_i=827.64\times 3.30+51.7\times 2.76\times 2.63$

$L_i=2731.212+375.27=3106.48 Js$

Final Moment inertia

$I_f=0.5Mr^2+mr^2$

$I_f=(152\cdot 0.5+51.7)\cdot (2.63)^2=1185.243 kg-m^2$

final angular momentum

$L_f=I_f\omega _f$

Conserving angular momentum

$L_i=L_f$

$3106.48=1408.97\times \omega _f$

$\omega _f=2.62 rad/s$

4 0

3 years ago

A 2 kg blob of putty moving at 3 m/s slams into a 3 kg blob of putty at rest. Which blob has the most momentum prior to the coll

Ray Of Light [21]

initial momentum of 2 kg blob is given as

$P_1 = m_1v_1$

here we have

$m_1 = 2kg$

$v_1 = 3m/s$

$P_1 = 3(2) = 6 kg m/s$

initial momentum of 3 kg blob is given as

$P_2 = m_2v_2$

here we have

$m_2 = 3kg$

$v_2 = 0m/s$

$P_2 = 3(0) = 0 kg m/s$

So initially 2 kg Blob has most momentum before they collide

5 0

3 years ago

4. Think back to Coulomb's Law. Two coins with identical charges are placed on a lab table 1.35 m apart.

FinnZ [79.3K]

A) To calculate the charge of each coin, we must apply the expression of the Coulomb's Law:

F=K(q1xq2)/r²

F: The magnitud of the force between the charges. (F=2.0 N).
K: Constant of proporcionality of the Coulomb's Law (K=9x10^9 Nxm²/C²).
q1 and q2: Electrical charges.
r: The distance between the charges (r=1.35 m).

We have the values of F, K and r, so we can calculate q1xq2, because both<span> coins have identical charges:
</span>
q1xq2=(r²xF)/K
q1xq2=(1.35 m)²(2.0 N)/9x10^9 Nxm²/C²
q1xq2=3x10^-10 C
q1=q2=(<span>3x10^-10 C)/2

</span>Then, the charge of each coin, is:
<span>
q1=1.5x</span><span>10^-10 C

</span>q2=1.5x10^-10 C

B) <span>Would the force be classified as a force of attraction or repulsion?
</span>
It is a force of repulsion, because both coins have identical charges and both are postive. In others words, when two bodies have identical charges (positive charges or negative charges), the force is of repulsion.

5 0

2 years ago

Read 2 more answers

A flashlight beam makes an angle of 60 degrees with the surface of the water before it enters the water. in the water what angle

alexira [117]

<h3><u>Answer</u>;</h3>

= 22°

<h3><u>Explanation</u>;</h3>

According to Snell's law, the ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant. The constant value is called the refractive index of the second medium with respect to the first.

Therefore; Sin i/Sin r = η

In this case; Angle of incidence = 90° -60° =30°, angle of refraction =? and η = 1.33

Thus;

Sin 30 / Sin r = 1.33

Sin r = Sin 30°/1.33

= 0.3759

r = Sin^-1 0.3759

= 22.08

<u>≈ 22°</u>

3 0

3 years ago

The drag force pushes opposite your motion as you ride a bicycle. If you double your speed, what happens to the magnitude of the

PIT_PIT [208]

The drag force is directly proportional to the square of the velocity of motion of the object. So as the speed is doubled, the magnitude of drag force will get quadrupled.

<u>Explanation: </u>

Drag force is the opposing or resisting force acting on any object by the medium in which it is moving. So in this case, you are riding a bicycle, thus the medium can be considered as air.

The formula for calculating drag force is as below:

$\text {Drag force }=\frac{1}{2} \rho v^{2} C_{D} A$

Here, ρ is the density of the air molecules, v is the velocity or speed of the bicycle, CD is the drag coefficient and A is the area of the bicycle.

In the above equation, only the term velocity will be a varying quantity with respect to time and other quantities will remain constant throughout the single situation of riding of bicycle.

So, the equation can be,

$\text { Drag force }=k v^{2}$

Where ,

$k=\frac{1}{2} \rho C_{D} A$ (constant for this whole condition)

Now given the speed of bicycle increased from v to 2v, so the initial drag force will be

$N_{i}=k v^{2}$

After increase in speed, the final drag force will be

$N_{f}=k(2 v)^{2}$

$N_{f}=4 k v^{2}=4 N_{i}$

Thus, if the speed of the bicycle is doubled, the drag force will get increased by four times.

4 0

3 years ago