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sweet [91]
3 years ago
7

A car's fuel economy is 30 mpg. a warning light is displayed if the remaining distance that can be driven before the car runs ou

t of gas drops below 50 miles. initially the car has 10 gallons of gas. how many gallons, g, can be used before the warning light comes on?
Physics
2 answers:
Alchen [17]3 years ago
7 0
So 10 gallons of gas would let you travel 300 Miles. 

x gallons = 50 Miles

10 : 300 :: x : 50 

x = 500/300

x = 1.66667 gallons. 

So, the car would run 10 - 1.6666 gallons = 8.33 gallons.

After that, the warning light turns ON! 

Hope this helps!!
shutvik [7]3 years ago
6 0

Answer:

8.33 gallons

Explanation:

Given: A car's fuel economy is 30 mpg. a warning light is displayed if the remaining distance that can be driven before the car runs out of gas drops below 50 miles.

To Find: initially the car has 10 gallons of gas. how many gallons, g, can be used before the warning light comes on.

Solution:

Total fuel in car=10\text{g}

fuel economy of car =30\text{mpg}

fuel required to ride 30 mile =1\text{g}

fuel required to ride 1 mile =\frac{1}{30}\text{g}

Now,

Fuel required to ride 50 miles =50\times\frac{1}{30}\text{g}=\frac{5}{3}\text{g}

fuel left in car when warning light is displayed =\text{Fuel required to ride 50 miles left}

                                                                        =\frac{5}{3}\text{g}

Amount of fuel that can be consumed before warning light dislays

\text{Total fuel}-\text{fuel left when warning light displays}

            10-\frac{5}{3}

            \frac{25}{3}\text{g}        

            8.33\text{g}

8.33 gallons fuel can be used before warning light comes on

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A hockey puck with a mass of 0.159 kg slides over the ice. The puck initially slides with a speed of 4.75 m/s, but it comes to a
Anton [14]

Answer:

The energy dissipated as the puck slides over the rough patch is 1.355 J

Explanation:

Given;

mass of the hockey puck, m = 0.159 kg

initial speed of the puck, u = 4.75 m/s

final speed of the puck, v = 2.35 m/s

The energy dissipated as the puck slides over the rough patch is given by;

ΔE = ¹/₂m(v² - u²)

ΔE = ¹/₂ x 0.159 (2.35² - 4.75²)

ΔE = -1.355 J

the lost energy is 1.355 J

Therefore, the energy dissipated as the puck slides over the rough patch is 1.355 J

5 0
3 years ago
You dip a wire loop into soapy water (n = 1.33) and hold it up vertically to look at the soap film in white light. The soap film
Gnesinka [82]

Answer:

the thickness of the soap film is 127.82 nm or 130 nm

Explanation:

Given the data in the question;

n = 1.33

λ_t = 680 nm = 680 × 10⁻⁹ m

m = 1 and β = 0

When we see the red fringe, its a point of maximum reflection

hence, for interference with a thin soap film, we say;

2 × n × d × cos( β ) = ( m - 0.5) × λ_t

so we substitute in our given values;

2 × 1.33 × d × cos( 0 ) = ( 1 - 0.5) × ( 680 × 10⁻⁹ )

2.66 × cos( 0 ) × d = 0.5 × ( 680 × 10⁻⁹ )

2.66 × 1 × d = 3.4 × 10⁻⁷

d = ( 3.4 × 10⁻⁷ ) / 2.66

d = 127.82 × 10⁻⁹ m

d = 127.82 nm ≈ 130 nm

Therefore, the thickness of the soap film is 127.82 nm or 130 nm

4 0
3 years ago
Two solid spheres are made from the same material, but one has twice the diameter of the other. Which sphere will have the great
aivan3 [116]

Answer:

It will be the same for both

Explanation:

from this question we have one similarity between these two spheres.

- they are both made from the same material,

The difference between both spheres is that:

- one of the spheres has its diameter to be twice as large as that of the other one.

We are to say the sphere with the greater bulk modulus.

If the material is the same thenthe Bulk modulus is also the same. It is not dependent on the material since it is a constant for that materia

Therefore the correct answer is:

It will be the same for both spheres.

5 0
3 years ago
Why don't we use more renewable energy sources​
sattari [20]

Answer:

Renewable energy sources 1.Cost more then nonrenewable resources and 2.Are so far not as reliable as nonrenewable energy.

Explanation:

4 0
3 years ago
Read 2 more answers
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kakasveta [241]

Answer:

The correct answer is  a. Both are the same

Explanation:

For this calculation we must use the gravitational attraction equation

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Where M will use the mass of the Earth, m the mass of the girl and r is the distance of the girl to the center of the earth that we consider spherical

To better visualize things, let's repair the equation a little

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The amount in parentheses called acceleration of gravity, entered the force called peos

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When analyzing this equation we see that the variation in the weight of the girl depends on the distance, which is the radius of the earth plus the height where the girl is

    r = Re + h

    Re = 6.37 10⁶ m

    r² = (Re + h)²

    r² = Re² (1 + h / Re)²

Let's replace

    W = m (GM / Re²)   (1+ h / Re)⁻²

    W = m g   (1+ h / Re)⁻²

This is the exact expression for weight change with height, but let's look at its values ​​for some reasonable heights h = 6300 m (very high mountain)

     h / Re = 10 ⁻³

     (1+ h / Re)⁻² = 0.999⁻²

Therefore, the negligible weight reduction, therefore, for practical purposes the weight does not change with the height of the mountain on Earth

The correct answer is a

4 0
3 years ago
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