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Marianna [84]
3 years ago
12

A certain fuel-efficient hybrid car gets gasoline mileage of 55.0 mpg (miles per gallon). (a) If you are driving this car in Eur

ope and want to compare its mileage with that of other European cars, express this mileage in km/L (L = liter). Use the conversion factors in Appendix E. (b) If this car’s gas tank holds 45 L, how many tanks of gas will you use to drive 1500 km?
Physics
1 answer:
ololo11 [35]3 years ago
6 0

We will start by defining the units and their respective equivalences between the proposed measurement systems

1km = 0.6214mi

1gallon = 3.788 litres

PART A ) The mileage of the car is 55mpg (Miles per gallon)

55mpg = 55(\frac{miles}{gallon}) (\frac{1km}{0.6214miles})(\frac{1gallon}{3.788L})

55mpg = 23.4km/L

Therefore the mileage of the car is 23.4km/L

PART B ) The mileage of the car means that the car travels 23.4km and consumes 1 liter of fuel. Then

1L = 23.4km

For the car to travel 1500km the amount of fuel would be,

1500km= (1500km)(\frac{1L}{23.4km})

1500km = 64.1L

But 1 gas tank can only hold 45Liters of fuel, then the number of tank required would be

\text{Number of tanks required} = \frac{64.1L}{45L}

\text{Number of tanks required} = 1.4tanks

Thus the number of tanks of gas required to drive 1500km is 1.4

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1) A train accelerates from 36 km/hr to 54 km/hr in 10<br> s. Find acceleration?
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Answer: Given:

Initial velocity= 36km/h=36x5/18=10m/s

Final velocity =54km/h=54x5/18=15m/s

Time =10sec

Acceleration = v-u/ t

=15-10/10=5/10=1/2=0.5 m/s2

Distance =s=?

From second equation of motion:

S=ut +1/2 at^2

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=100+25

=125m

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Hope it helps you

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3 years ago
Precision is how close a series of measurements are to ______
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2 years ago
The speed of water flowing through a hose increases from 2.05 m/s to 31.4 m/s as it goes through the nozzle. What is the pressur
Nimfa-mama [501]

The pressure in the hose as the speed of water changes from 2.05 m/s to 31.4 m/s as it goes through the nozzle is 5.92 × 10⁵ N/m².

Given:

The flow of water through the hose initially, v₁ = 2.05 m/s

The flow of water through the hose initially, v₂ = 31.4 m/s

Calculation:

From Bernoulli's equation we have:

P₁ + 1/2 ρv₁² + ρgh₁ = P₂ + 1/2 ρv₂² + ρgh₂

where P₁ is atmospheric pressure

           P₂ is the pressure in the hose

           ρ is the density of the fluid

           h₁ is the initial height

           h₂ is the final height

           v₁ is the initial velocity of the fluid

           v₂ is the final velocity of the fluid  and

           g is the acceleration due to gravity

Re-arranging the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (h₁-h₂)

Applying values in the above equation we get:

P₂ = P₁ + 1/2 ρ(v₁²-v₂²) + ρg (0)

    = (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [(31.4m/s)²-(2.05 m/s)²]

    = (1.01 × 10⁵ Pa)+ 1/2 (10³ g/m³) [981.7575]

    = (1.01 × 10⁵ Pa)+ (4.91 × 10⁵ Pa)

    = 5.92 × 10⁵ Pa

    = 5.92 × 10⁵ N/m²

Therefore, the pressure in the hose is 5.92 × 10⁵ N/m².

Learn more about Bernoulli's equation here:

<u>brainly.com/question/9506577</u>

#SPJ4

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