Ask question

Ask question

All categories

3 years ago

15

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.

You might be interested in

Name the stages of a volcano and name the three different types of volcano mountains caused by the lava buildup.

Rudik [331]

Answer:

Active—A volcano is active if it is erupting, or may erupt soon. ...

Dormant—A dormant volcano is one that may have erupted before, but it is no longer erupting. ...

Extinct—An extinct volcano is not erupting and will never erupt again.

Explanation:

5 0

2 years ago

Read 2 more answers

The volume flow rate of the water supplied by a well is 2.0×10−4m3/s.The well is 40.0 m deep. (a) What is the power output of th

MaRussiya [10]

Answer:

a). $P=78.4W$

b). $P=392kPa$

c.) It must be at the bottom

Explanation:

Given:

Volume flow $V_f=2.0x10^{-4}m^3/s$

Well depp $h=40.m$

a.

The power output of the pum

$W=F*d$

$F=m*g$

$m=p*V=1000kg/m^3*2.0x10^{-4}m^3}=0.2Kg$

$W=m*g*d=0.2kg*9.8m/s^2*40.0m=78.4kg*m^2/s^2$

$W=78.4J$

$P=\frac{W}{t}=\frac{78.4J}{1s}=78.4W$

b.

The pressure of difference the pum

Δ $P=p*g*y'$

Δ $P=1000kg/m^3*9.8m/s^2*40.0m=392x10^3Pa$

$P=392kPa$

c.

It must be at the bottom since the pressure difference is greater than atmospheric pressure, so it wouldn't be able to lift the water all the way

4 0

3 years ago

The rocket's acceleration has components \(a_{x}(t)= \alpha t^{2}\) and \(a_{y}(t)= \beta - \gamma t\), where \(\alpha = 2.50 {\

lbvjy [14]

it is just a matter of integration and using initial conditions since in general dv/dt = a it implies v = integral a dt
v(t)_x = integral a_{x}(t) dt = alpha t^3/3 + c the integration constant c can be found out since we know v(t)_x at t =0 is v_{0x} so substitute this in the equation to get v(t)_x = alpha t^3 / 3 + v_{0x}
similarly v(t)_y = integral a_{y}(t) dt = integral beta - gamma t dt = beta t - gamma t^2 / 2 + c this constant c use at t = 0 v(t)_y = v_{0y} v(t)_y = beta t - gamma t^2 / 2 + v_{0y}
so the velocity vector as a function of time vec{v}(t) in terms of components as[ alpha t^3 / 3 + v_{0x} , beta t - gamma t^2 / 2 + v_{0y} ]
similarly you should integrate to find position vector since dr/dt = v r = integral of v dt
r(t)_x = alpha t^4 / 12 + + v_{0x}t + c let us assume the initial position vector is at origin so x and y initial position vector is zero and hence c = 0 in both cases
r(t)_y = beta t^2/2 - gamma t^3/6 + v_{0y} t + c here c = 0 since it is at 0 when t = 0 we assume
r(t)_vec = [ r(t)_x , r(t)_y ] = [ alpha t^4 / 12 + + v_{0x}t , beta t^2/2 - gamma t^3/6 + v_{0y} t ]

5 0

3 years ago

The area of the piston to the master cylinder in a hydraulic braking system of a car is 0.4 square inches. If a force of 6.4 lb

Anit [1.1K]

Answer:

The force applied on one wheel during braking = 6.8 lb

Explanation:

Area of the piston (A) = 0.4 $in^{2}$

Force applied on the piston(F) = 6.4 lb

Pressure on the piston (P) = $\frac{F}{A}$

⇒ P = $\frac{6.4}{0.4}$

⇒ P = 16 $\frac{lb}{in^{2} }$

This is the pressure inside the cylinder.

Let force applied on the brake pad = $F_{1}$

Area of the brake pad ( $A_{1}$ )= 1.7 $in^{2}$

Thus the pressure on the brake pad ( $P_{1}$ ) = $\frac{F_{1} }{A_{1} }$

When brake is applied on the vehicle the pressure on the piston is equal to pressure on the brake pad.

⇒ P = $P_{1}$

⇒ 16 = $\frac{F_{1} }{A_{1} }$

⇒ $F_{1}$ = 16 × $A_{1}$

Put the value of $A_{1}$ we get

⇒ $F_{1}$ = 16 × 1.7

⇒ $F_{1}$ = 27.2 lb

This the total force applied during braking.

The force applied on one wheel = $\frac{F_{1} }{4}$ = $\frac{27.2}{4}$ = 6.8 lb

⇒ The force applied on one wheel during braking.

7 0

2 years ago

If you have 5.112 grams of ammonia (NH3), then how many molecules of ammonia do you have?

Dominik [7]

I hope it helps! good luck in chem!

7 0

2 years ago