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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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Paraphin [41]

Answer : The average speed of the sprinter is, 34.95 Km/hr

Solution :

Average velocity : It is defined as the distance traveled by the time taken.

Formula used for average velocity :

v_{av}=\frac{d}{t}

where,

v_{av} = average velocity

d = distance traveled = 200 m

t = time taken = 20.6 s

Now put all the given values in the above formula, we get the average velocity of the sprinter.

v_{av}=\frac{200m}{20.6s}\times \frac{3600}{1000}=34.95Km/hr

conversion :

(1 Km = 1000m)

(1 hr = 3600 s)

Therefore, the average speed of the sprinter is, 34.95 Km/hr

8 0
2 years ago
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Suppose that a public address system emits sound uniformly in all directions and that there are no reflections . The intensity a
Marina86 [1]

Answer:

I_{2}=2.39*10^{-5} W/m^{2}

Explanation:

The definition of the intensity in terms of power is given by:

I=\frac{P}{A}

Where:

  • P is the power
  • A is the area

If the sound emits uniformly in all directions and that there are no reflections, we can assume the geometry of the wave sound is spherical.

Let's recall the area of a sphere is A = 4\pi R^{2}

To the first location we have:

I_{1}=3*10^{-4} W/m^{2}=\frac{P}{4\pi 22^{2}}

and to the second location we have:

I_{2}=\frac{P}{4\pi 78^{2}}

Now, we can divide each intensity to find the second intensity.

\frac{I_{2}}{I_{1}}=\frac{\frac{P}{4\pi 78^{2}}}{\frac{P}{4\pi 22^{2}}}

I_{2}=I_{1}* \frac{22^{2}}{78^{2}}

I_{2}=3*10^{-4}\frac{22^{2}}{78^{2}}

I_{2}=2.39*10^{-5} W/m^{2}

I hope it helps you!

     

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3 years ago
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A traves de una manguera de 1 in de diámetro fluye gasolina con una velocidad media de 5ft/s ¿cuál es el gasto?
jonny [76]

Answer:

El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

Explanation:

El gasto es el flujo volumétrico de gasolina (Q), medido en pies cúbicos por segundo, que sale de la manguera. Asumiendo que la velocidad de salida es constante, tenemos que el gasto a través de la manguera es:

Q = \frac{\pi}{4}\cdot D^{2}\cdot v (1)

Donde:

D - Diámetro de la manguera, medido en pies.

v - Velocidad medida de salida, medida en pies por segundo.

Si sabemos que D = \frac{1}{12}\,ft y v = 5\,\frac{ft}{s }, entonces el gasto de gasolina es:

Q = \frac{\pi}{4}\cdot \left(\frac{1}{12}\,ft \right)^{2} \cdot \left(5\,\frac{ft}{s} \right)

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El gasto de gasto es de aproximadamente 0.0273 pies cúbicos por segundo.

6 0
3 years ago
The smallest separation resolvable by a microscope is of the order of magnitude of the wavelength used. What energy electrons wo
hjlf

Answer: a) for 150 Angstroms 6.63 *10^-3 eV; b) for 5 Angstroms 6.02 eV

Explanation: To solve this problem we have to use the relationship given by De Broglie as:

λ =p/h where p is the momentum and h the Planck constant

if we consider the energy given by acceleration tube for the electrons given by: E: e ΔV so is equal to kinetic energy of electrons p^2/2m

Finally we have:

eΔV=p^2/2m= h^2/(2*m*λ^2)

replacing we obtained the above values.

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