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GarryVolchara [31]
4 years ago
9

Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of poun

ds), and let y be the miles per gallon (mpg)
Engineering
1 answer:
RoseWind [281]4 years ago
6 0

Answer:

Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of pounds), and let y be the miles per gallon (mpg)

Explanation:

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List at least two things that scientists and explorers have in common.
Tresset [83]
• educated
•very curious
•discover new things
•Investigate
•search unknown
8 0
3 years ago
Read 2 more answers
You buy a 75-W lightbulb in Europe, where electricity is delivered to homes at 240V. If you use the lightbulb in the United Stat
givi [52]

Answer:

The bulb will be \frac{1}{4} times as bright as it is in Europe.

Explanation:

Data provided in the question:

Power of bulb in Europe = 75 W

Voltage provided in Europe = 240 V

Voltage provided in United Stated = 120 V

Now,

We know,

Power = Voltage² ÷ Resistance

Therefore,

75 = 240² ÷ Resistance

or

Resistance = 240² ÷ 75

or

Resistance = 768 Ω

Therefore,

Power in United States = Voltage² ÷ Resistance

= 120² ÷ 768

= 18.75 W

Therefore,

Ratio of powers

\frac{\text{Power in united states}}{\text{Power in Europe}}=\frac{18.75}{75}

= \frac{1}{4}

Hence,

The bulb will be \frac{1}{4} times as bright as it is in Europe.

7 0
3 years ago
What substance does light travel through before putting water in the cup
torisob [31]
Bend surface in water! Hopefully this helps, I looked it up!

6 0
3 years ago
Carbon dioxide used as a natural refrigerant flows through a cooler at 10 MPa, which is supercritical, so no condensation occurs
kupik [55]

Answer:

The answer which is a calculation can be found as an attached document

Explanation:

5 0
3 years ago
Read 2 more answers
The Molybdenum with an atomic radius 0.1363 nm and atomic weight 95.95 has a BCC unit cell structure. Calculate its theoretical
Nookie1986 [14]

Solution:

Given :

atomic radius, r = 0.1363nm = 0.1363×10⁻⁹m

atomic wieght, M = 95.96

Cell structure is BCC (Body Centred Cubic)

For BCC, we know that no. of atoms per unit cell, z = 2

and atomic radius, r =\frac{a\sqrt{3} }{4}

so, a = \frac{4r}{\sqrt{3}}

m = mass of each atom in a unit cell

mass of an atom = \frac{M}{N_{A} },

where, N_{A} is Avagadro Number = 6.02×10^{23}

volume of unit cell = a^{3}

density, ρ = \frac{mass of unit cell}{volume of unit cell}

density, ρ = \frac{z\times M}{a^{3}\times N_{A}}

ρ = \frac{2\times 95.95}{(\frac{4\times 0.1363\times 10^{-9}}{\sqrt{3}})^{3}\times (6.23\times 10^{23})}

ρ = 10.215gm/cm^{3}

5 0
4 years ago
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