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vazorg [7]
3 years ago
14

Which of the following is not a requirement of a practical fuel? A. It must occur in abundance in nature or be easy to produce.

B. It must be made up of elements that combine easily with oxygen. C. It must contain a large amount of stored energy. D. It must be in the solid state to be used.
Physics
1 answer:
Masteriza [31]3 years ago
7 0

Answer: Option D: It must be in the solid state to be used.

A material that is used to provide energy is for work is known as fuel. For practical use, it is important that the fuel must occur in abundance or can be easily produced. It should be made of elements that can easily combine with oxygen. This is required for burning. It must produce large amount of energy. So, it must contain large amount of stored energy. It is useful only when a small quantity can produce large amount of energy. It can be in any state, solid liquid or gas.

Therefore, it is not important that it must be in the solid state to be used.

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Use Kepler’s third law and the orbital motion of Earth to determine the mass of the Sun. The average distance between Earth and
dexar [7]

Kepler’s
third law formula: T^2=4pi^2*r^3/(GM)

We’re trying to find M, so:

M=4pi^2*r^3/(G*T^2)

M=4pi^2*(1.496
× 10^11 m)^3/((6.674× 10^-11N*m^2/kg^2)*(365.26days)^2)

M=1.48× 10^40(m^3)/((N*m^2/kg^2)*days^2))

Let’s work
with the units:

(m^3)/((N*m^2/kg^2)*days^2))=

=(m^3*kg^2)/(N*m^2*days^2)

=(m*kg^2)/(N*days^2)

=(m*kg^2)/((kg*m/s^2)*days^2)

=(kg)/(days^2/s^2)

=(kg*s^2)/(days^2)

So:

M=1.48× 10^40(kg*s^2)/(days^2)

Now we need to convert days to seconds in order to cancel
them:

1 day=24 hours=24*60minutes=24*60*60s=86400s

M=1.48× 10^40(kg*s^2)/((86400s)^2)

M=1.48× 10^40(kg*s^2)/(
86400^2*s^2)

M=1.48× 10^40kg/86400^2

M=1.98x10^30kg

The
closest answer is 1.99
× 10^30

(it may vary
a little with rounding – the difference is less than 1%)


8 0
3 years ago
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2. Which of the following is NOT true of work?
salantis [7]

Answer:

D

Explanation:

Work is not a vector but it is a scalar

3 0
2 years ago
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An engineering team has come to the stage in the engineering design
SIZIF [17.4K]

Answer: D

Explanation: read the answers above and compare to your test

6 0
3 years ago
Two long, parallel wires separated by 3.50 cm carry currents in opposite directions. The current in one wire is 1.55 A, and the
vaieri [72.5K]

Answer:

Therefore,

The magnitude of the force per unit length that one wire exerts on the other is

\dfrac{F}{l}=2.79\times 10^{-5}\ N/m

Explanation:

Given:

Two long, parallel wires separated by a distance,

d = 3.50 cm = 0.035 meter

Currents,

I_{1}=1.55\ A\\I_{2}=3.15\ A

To Find:

Magnitude of the force per unit length that one wire exerts on the other,

\dfrac{F}{l}=?

Solution:

Magnitude of the force per unit length on each of @ parallel wires seperated by the distance d and carrying currents I₁ and I₂ is given by,

\dfrac{F}{l}=\dfrac{\mu_{0}\times I_{1}\times I_{2}}{2\pi\times d}

where,

\mu_{0}=permeability\ of\ free\ space =4\pi\times 10^{-7}

Substituting the values we get

\dfrac{F}{l}=\dfrac{4\pi\times 10^{-7}\times 1.55\times 3.15}{2\pi\times 0.035}

\dfrac{F}{l}=2.79\times 10^{-5}\ N/m

Therefore,

The magnitude of the force per unit length that one wire exerts on the other is

\dfrac{F}{l}=2.79\times 10^{-5}\ N/m

7 0
3 years ago
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The speed of light is the fastest in which medium
wlad13 [49]

In vacuum, going at 2.99×10^8 m/s.

3 0
3 years ago
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