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saul85 [17]
2 years ago
8

Estimate the number of gallons of gasoline consumed by the total of all automobile drivers in the U.S., per year. Suppose that t

here are about 3 × 10^8 people in the United States, approximately half of the them have cars, each car drives an average of 12,000 mi per year, and consumes a gallon of gasoline for each 20 mi?
Physics
2 answers:
Daniel [21]2 years ago
8 0

Answer:

G = 9 \times 10^{10} gallons

Explanation:

Total number of people in US is  N_p = 3 \times 10^8

Only half of them have cars. So, Number of cars in US

N_C = \frac{N_p}{2} \\\\N_C = \frac{3 \times 10^8}{2} \\\\N_C = 1.5 \times 10^8

Each cars drives 12000 miles, so total distance travelled by all of these cars combined

D = N_C \times 12000\\\\D = 1.5 \times 10^8 \times 12000\\\\D = 1.8 \times 10^{12} miles

To travel 20 miles, we need 1 gallon of gasoline

Amount of gasoline required to travel one mile is  \frac{1}{20} gallons

Amount of gasoline required to cover the distance 'D'.

G = D \times \frac{1}{20} \\\\G = 1.8 \times 10^{12} \times \frac{1}{20} \\\\G = 9 \times 10^{10} gallons

Amount of gasoline required to cover the distance 'D'.

G = D \times \frac{1}{20} \\\\G = 1.8 \times 10^{12} \times \frac{1}{20} \\\\G = 9 \times 10^{10} gallons

vaieri [72.5K]2 years ago
3 0
3 x 108 is roughly 300 people. Half of them have cars. Half of 300 = 150. 150 x 12000 = 1,800,000 miles driven. Each car gets 20mpg. Solve for the # of gallons consumed.
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The Coulomb force is equal to the constant k times the product of charge one and charge two over radius.

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6 0
2 years ago
Two 5.0 mm × 5.0 mm electrodes are held 0.10 mm apart and are attached to 7.5 V battery. Without disconnecting the battery, a 0.
Musya8 [376]

Answer:

A) V = 7.5 V

B) E = 75,000 V/m

C) Q = 16.6 pC

D) V = 7.5 V

E) E = 24,000 V/m

F) Q = 52 pC

Explanation:

Given:

- The Area of plate A = ( 5 x 5 ) mm^2

- The distance between plates d = 0.10 mm

- The thickness of Mylar added t = 0.10 mm

- Voltage supplied by battery V = 7.5 V

Solution:

A) What is the capacitor's potential difference before the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

B) What is the capacitor's electric field before the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

                                E = V / k*d

k = 1 (air)                  E = 7.5 / 0.10*10^-3

                                E = 75,000 V/m

C) What is the capacitor's charge Q before the Mylar is inserted?

                                C = k*A*ε / d

k = 1 (air)                   C = ( 0.005^2 * 8.85*10^-12 ) / 0.0001

                                C = 2.213 pF

                                Q = C*V

                                Q = 7.5*(2.213)

                                Q = 16.6 pC

D) What is the capacitor's potential difference after the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

E) What is the capacitor's electric field after the Mylar is inserted?    

- The Electric Field E between the capacitor plates is given by:

                                E = V / k*d

k = 3.13                     E = 7.5 / (3.13)0.10*10^-3

                                E = 24,000 V/m              

F) What is the capacitor's charge after the Mylar is inserted?      

                                C = k*A*ε / d

k = 3.13                    C = 3.13*( 0.005^2 * 8.85*10^-12 ) / 0.0001

                                C = 6.927 pF

                                Q = C*V

                                Q = 7.5*(6.927)

                                Q = 52 pC                                      

6 0
2 years ago
Keisha looks out the window from a tall building at her friend Monique standing on the ground, 8.3 m away from the side of the b
Salsk061 [2.6K]

Answer:

Explanation:

GIVEN DATA:

Distance between keisha and her friend 8.3 m

angle made by keisha toside building 30 degree

height of her friend monique is 1.5 m

from the figure

\Delta ACB

tan 30 = \frac{8.3}{h}

h= \frac{8.3}{tan 30} = 14.376 m

therefore

height of keisha is = h  + 1.5 m

                               = 14.376 + 1.5

= 15.876 \simeq 16 m

therefore option c is correct

5 0
2 years ago
At t = 0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 30.0rad/s2 until a ci
kozerog [31]

Answer:

θ=108rad

t =10.29seconds

α=-8.17rad/s²

Explanation:

Given that

At t=0, Wo=24rad/sec

Constant angular acceleration =30rad/s²

At t=2, θ=432rad as it try to stop because the circuit break

Angular motion

W=Wo+αt

θ=Wot+1/2αt²

W²=Wo²+2αθ

We need to find θ between 0sec to 2sec when the wheel stop

a. θ=Wot+1/2αt²

θ=24×2+1/2×30×2²

θ=48+60

θ=108rad.

b. W=Wo+αt

W=24+30×2

W=84rad/s

This is the final angular velocity which is the initial angular velocity when the wheel starts to decelerate.

Wo=84rad/sec

W=0rad/s, because the wheel stop at θ=432rad

Using W²=Wo²+2αθ

0²=84²+2×α×432

-84²=864α

α=-8.17rad/s²

It is negative because it is decelerating

Now, time taken for the wheel to stop

W=Wo+αt

0=84-8.17t

-84=-8.17t

Then t =10.29seconds.

a. θ=108rad

b. t =10.29seconds

c. α=-8.17rad/s²

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2 years ago
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