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Veronika [31]
4 years ago
5

Diesel engines burn as much as 30% less fuel than gasoline engines of comparable size, as well as emitting far less carbon dioxi

de gas and far fewer of the other gasses that have been implicated in global warming.
(A) of comparable size, as well as emitting far less carbon dioxide gas and far fewer of the other gasses that have

(B) of comparable size, as well as emit far less carbon dioxide gas and far fewer of the other gasses having

(C) of comparable size, and also they emit far fewer carbon dioxide and other gasses that have

(D) that have a comparable size, and also they emit far fewer of the other gasses having

(E) that have a comparable size, as well as emitting far fewer of the other gasses having
Physics
1 answer:
Gennadij [26K]4 years ago
6 0

Answer:

Explanation:

<em>Diesel engines burn as much as 30% less fuel than gasoline engines of comparable size, as well as emitting far less carbon dioxide gas and far fewer of the other gasses that have been implicated in global warming. </em>

<em> </em>

<em> </em>

<em>(A) of comparable size, as well as emitting far less carbon dioxide gas and far fewer of the other gasses that have </em>

<em> </em>

<em>(B) of comparable size, as well as emit far less carbon dioxide gas and far fewer of the other gasses having </em>

<em> </em>

<em>(C) of comparable size, and also they emit far fewer carbon dioxide and other gasses that have </em>

<em> </em>

<em>(D) that have a comparable size, and also they emit far fewer of the other gasses having </em>

<em> </em>

<em>(E) that have a comparable size, as well as emitting far fewer of the other gasses having</em>

<em>A is the most appropriate statement above. Diesel engine consume less energy and release fewer gases than the gasoline engines .</em>

<em>Both engine(Diesel and gasoline) converts chemical energy to electrical energy.</em>

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Answer:

a.F=7.83\times 10^{-51} N

b.Attractive

Explanation:

We are given that

F=\frac{GM_1M_2}{R^2}

G=6.67\times 10^{-11} Nm^2/kg^2

Mass of an electron,M_1=9.11\times 10^{-31} kg

Mass of proton,M_2=1836\times 9.11\times 10^{-31} kg

Distance between electron and proton,R=3.602nm=3.602\times 10^{-9} m

1nm=10^{-9} m

a.Substitute the values then  we get

F=\frac{6.67\times 10^{-11}\times 9.11\times 10^{-31}\times 1836\times 9.11\times 10^{-31}}{(3.602\times 10^{-9})^2}

F=7.83\times 10^{-51} N

b.We know that like charges repel to each other and unlike charges attract to each other.

Proton and electron are unlike charges therefore, the force between proton and electron is attractive.

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3 years ago
If the mass of an object is 10 kg and the
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B. -40 kgm/s is the answer

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In the qualifying round of the 50-yard freestyle in the sectional swimming championship, Dugan got an early lead by finishing th
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Answer:

V_{average}=  2.471  yards/s

Explanation:

We can calculate the average velocity thus:

V_{average} = \frac{D_{1}+D_{2} }{t_{1}+t_{2}}

Where

              D_{1} : First 25 yards

              t_{1}: Time to travel  D_{1}

              D_{2} : Seconds 25 yards

              t_{2}: Time to travel  D_{2}

Thus

V_{average} = \frac{25 + 25}{10.01 + 10.22}

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The flow of ____ forms the physical basis for electric charge.
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Answer:

electrons

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electrons are the only subatomic particle that flow

7 0
3 years ago
At some instant and location, the electric field associated with an electromagnetic wave in vacuum has the strength 70.5 V/m. Fi
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Answer:

The magnetic field strength, the energy density, and the power flow per unit area are 2.35\times10^{-7}\ T, 4.396\times10^{-8}\ J/m^3 and 13.18 W/m².

Explanation:

Given that,

Electromagnetic wave strength E= 70.5 V/m

(I). We need to calculate the magnetic field strength

Using formula of Electromagnetic wave strength

c= \dfrca{E}{B}

B=\dfrac{E}{c}

B=\dfrac{70.5 }{3\times10^{8}}

B=2.35\times10^{-7}\ T

(II). We need to calculate the energy density

Using formula of energy density

\mu_{total}=\mu_{E}+\mu_{B}

\mu_{total}=\dfrac{1}{2}\epsilon_{0}E^2+\dfrac{1}{2}\dfrac{B^2}{\mu_{0}}

\mu_{total}=\dfrac{1}{2}\times8.85\times10^{-12}\times(70.5)^2+\dfrac{1}{2}\times\dfrac{(2.35\times10^{-7})^2}{4\pi\times10^{-7}}

\mu_{total}=4.396\times10^{-8}\ J/m^3

(III). We need to calculate the power flow per unit area

Using formula of poynting vector

S=\dfrac{1}{\mu_{0}}EB

Put the value into the formula

S=\dfrac{1}{4\pi\times10^{-7}}\times70.5\times2.35\times10^{-7}

S=13.18\ W/m^2

Hence, The magnetic field strength, the energy density, and the power flow per unit area are 2.35\times10^{-7}\ T, 4.396\times10^{-8}\ J/m^3 and 13.18 W/m².

5 0
3 years ago
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