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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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Q = magnitude of charge at each of the three locations A, B and C = 2 x 10⁻⁶ C

r₁ = distance of charge at origin from charge at B = 50 - 0 = 50 cm = 0.50 m

r₂ = distance of charge at origin from charge at C = 100 - 0 = 100 cm = 1 m

F₁ = magnitude of force by charge at B on charge at origin

F₂ = magnitude of force by charge at C on charge at origin

Magnitude of force by charge at B on charge at origin

F_{1}=\frac{kQ^{2}}{r_{1}^{2}}

inserting the values

F_{1}=\frac{(9\times 10^{9})(2 \times 10^{-6})^{2}}{0.5^{2}}

F₁ = 0.144 N


Magnitude of force by charge at C on charge at origin

F_{2}=\frac{kQ^{2}}{r_{2}^{2}}

inserting the values

F_{2}=\frac{(9\times 10^{9})(2 \times 10^{-6})^{2}}{1^{2}}

F₂ = 0.036 N

Net force on the charge at the origin is given as

F = F₁ + F₂

F = 0.144 + 0.036

F = 0.18 N

from the diagram , direction of net force is towards left or negative x-direction.


3 0
2 years ago
A boy falls and hits his head with an impulse of 20N-s. On cement, the boy would hit for
ANTONII [103]

Answer:

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Explanation:

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5 0
3 years ago
The revival of the atomic concept was catalyzed by what experimental observation?
IgorC [24]
It seems that you have missed the necessary options for us to answer this question so I had to look for it. Anyway, here is the answer. The revival of the atomic concept was catalyzed by the experimental observation which is Constant <span>composition in compounds. Hope this helps.</span>
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3 years ago
a 25 newton force applied on an object moves it 50 meters. the angle between the force and displacement is 40.0°. what is the va
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Work = force × distance × cos(angle)

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3 0
3 years ago
A capacitor is connected across an ac source. Suppose the frequency of the source is doubled. What happens to the capacitive rea
fgiga [73]

The capacitive reactance is reduced by a factor of 2.

<h3>Calculation:</h3>

We know the capacitive reactance is given as,

Xc = \frac{1}{2\pi fC}

where,

Xc\\ = capacitive reactance

f = frequency

C = capacitance

It is given that frequency is doubled, i.e.,

f' = 2f

To find,

Xc =?

Xc' = \frac{1}{2\pi f'C}

      = \frac{1}{2\pi 2f C}

      = \frac{1}{2} (\frac{1}{2\pi fC} )\\

Xc' = \frac{1}{2} Xc

Therefore, the capacitive reactance is reduced by a factor of 2.

I understand the question you are looking for is this:

A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?

  1. The capacitive reactance is doubled.
  2. The capacitive reactance is traduced by a factor of 4.
  3. The capacitive reactance remains constant.
  4. The capacitive reactance is quadrupled.
  5. The capacitive reactance is reduced by a factor of 2.

Learn more about capacitive reactance here:

brainly.com/question/23427243

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