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3 years ago

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One gallon of gasoline in an automobiles engine produces on average 9.50 kg of carbon dioxide, which is a greenhouse gas; that i

s, it promotes the warming of earth's atmosphere. calculate the annual production of carbon dioxide in kilograms if there are exactly 40.0 million cars in the united states and each car covers a distance of 6990 mi at a consumption rate of 21.4 miles per gallon. enter your answer in scientific notation.

1 answer:

Musya8 [376]3 years ago

4 0

The annual production of carbon dioxide is 124121.49×10^{6}[/tex] kg.

First we calculate the fuel consumed by each car in a year

Fuel consumed=6990/21.4=326.63 gallon

Now we calculate the amount of fuel consumed by 40 million cars in a year

Fuel consumed=326.63*40*10^6=13065.42 million gallon,

Now we can calculate the annual production of carbon dioxide in the USA

CO2 production rate=9.50*13065.42=124121.49*10^6 kg

Therefore the annual production of carbon dioxide in USA is 124121.49×10^{6}[/tex] kg

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WINSTONCH [101]

Answer:

Final vertical velocity = -29m/s

Horizontal distance = 100m

Height = 20.41m

Explanation:

1. The vertical final velocity can be calculated thus: vy = vyo - gt

Where;

vy = vertical velocity (m/s)

vyo = vertical initial velocity (20m/s)

g = acceleration due to gravity (9.8m/s²)

t = time (5s)

Hence, vy = vyo - gt

vy = 20 - (9.8 × 5)

vy = 20 - 49

vy = -29m/s

2. x = V0 x t

Where;

x = horizontal distance (m)

Vo = initial velocity

t = time (s)

x = 20 × 5

x = 100m

3. Maximum height = (voy)²/2g

= 20²/ 2 × 9.8

= 400/19.6

= 20.41m

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3 years ago

A double nozzle lying in a horizontal x-y plane discharges water into the atmosphere at a rate of 0.5 m3 /s. Assume the water sp

Kisachek [45]

Answer:

The force is $F= 46.25kN$

Explanation:

The diagram for this question is shown on the first uploaded image

At Equilibrium the summation of the of force on the vertical axis is zero

i.e $\sum F_y =0$

=> $F_y sin \ 60^o =\rho Q (v_2 -v_1 cos \ 30^o)$

$v_2$ is the is the speed of water at the nozzle which can be mathematically evaluated as

$v_2 = \frac{R}{A_n}$

substituting $0.5m^3/s$ for R and $\frac{\pi}{4}(12*\frac{1m}{100} )^2$ for $A_n$

$v_2 = \frac{0.5}{\frac{\pi}{4} * (12*\frac{1}{100} )^2 }$

$= 44.23 m/s$

$v_1$ is the is the speed of water at the pipe which can be mathematically evaluated as

$v_1 = \frac{R}{A_p}$

substituting $0.5m^3/s$ for R and $\frac{\pi}{4}(30*\frac{1m}{100} )^2$ for $A_p$

$v_1 = \frac{0.5}{\frac{\pi}{4} * (30*\frac{1}{100} )^2 }$

$= 7.07 m/s$

$\rho$ is he density of water with value $\rho =1000 kg /m^3$

Substituting values into the equation above

$F_ysin 60^o = 1000 (0.5) (44.23 -7.07 cos 30)$

$= 21.99kN$

At Equilibrium the summation of the of force on the horizontal axis is zero

i.e $\sum F_x =0$

=> $F_y sin \ 30^o =\rho Q (v_2 -v_1 sin \ 30^o)$

Since The speed at both A and B nozzle are the same then $v_2$ remains the same

Substituting values

$F_x sin30^o =1000 (0.5) (44.23 - 7.07*sin30)$

=> $F_x = 40.69kN$

Hence the force acting on the flange bolts required to hold the nozzle in place is

$F = \sqrt{F_x^2 + F_y^2}$

$= \sqrt{40.69 ^2 + 21.99^2}$

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3 years ago

Read 2 more answers

a motorcycle is capable of accelerating at 5.1 m/s starting from rest how far can it travell in 1.5 sec

Katarina [22]

The answer is 21m because the motion is in one dimension with constant acceleration.

The initial velocity is 0, because it started from rest, the acceleration <span>ax</span> is <span>4.7<span>m<span>s2</span></span></span>, and the time t is <span>3.0s</span>

Plugging in our known values, we have

<span>Δx=<span>(0)</span><span>(3.0s)</span>+<span>12</span><span>(4.7<span>m<span>s2</span></span>)</span><span><span>(3.0s)</span>2</span>=<span>21<span>m</span></span></span>

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3 years ago

An archer pulls her bowstring back 0.376 m by exerting a force that increases uniformly from zero to 251 N. (a) What is the equi

Amanda [17]

Answer:

(A) 667.5 N/m

(B)

Explanation:

(A) Let the spring constant be k.

Using the formula F = kx

k = 251 / 0.376

K = 667.5 N/m

(B)

Work done

W = 0.5 × kx^2

W = 0.5 × 667.5 × 0.376 × 0.376

W = 47.2 J

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3 years ago

Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 80.5 N, Jill pulls with 81.7 N in the

Gnesinka [82]

Answer:

F = 233.52 N, θ' = 351.41º

Explanation:

In this exercise we must find the net force applied on the donkey.

For this we use Newton's second law, where we create a reference frame with the horizontal x axis

let's decompose the forces

Jack

= 80.5 N

Jill

cos 45 = F_{2x} / F₂2

sin 45 = F_{2y} / F₂2

F_{2x} = F₂ cos 45

F_{2y} = F₂ sin 45

F_{2x} = 81.7 cos 45 = 57.77 N

F_{2y} = 81.7 sin 45 = 57.77 N

Jane

cos (270 + 45) = F_{3x} / F₃3

sin 315 = F_{3y} / F₃

F_{3x} = 131 cos 315 = 92.63 N

F_{3y} = 131 sin 315 = -92.63 N

the force can be found in each axis

X axis

F_{x} = F_{1x} + F_{2x} + F_{3x}

F_{x} = 80.5 +57.77 + 92.63

F_{x} = 230.9 N

Axis y

F_{y} = F_{1y} + F_{2y} + F_{3y}

F_{y} = 0 + 57.77 -92.63

F_{y} = -34.86 N

we can give the result in two ways

a) F = (230.9 i ^ - 34.86 j ^) N

b) in the form of module and angle

we use the Pythagorean theorem

F = √(Fₓ² + F_{y}²

F = √(230.9² + 34.86²)

F = 233.52 N

let's use trigonometry for the angle

tan θ = $\frac{F_y}{F_x} }$

θ = tan⁻¹ (\frac{F_y}{F_x} })

θ = tan⁻¹ (-34.86 / 230.9)

θ = -8.59º

if we measure this angle from the positive side of the x-axis counterclockwise

θ' = 360 -θ

θ‘= 360- 8.59

θ' = 351.41º

5 0

3 years ago