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natka813 [3]
2 years ago
14

An internal combustion engine does 356 kJ of useful work using 946 kJ of thermal energy from the gasoline consumed in the engine

. What is the efficiency of this engine?
Physics
1 answer:
OlgaM077 [116]2 years ago
3 0

Answer:

The efficiency of this engine is 37.63 %.

Explanation:

Given;

useful output work done by the combustion engine, = 356 kJ

input work, = 946 kJ

The efficiency of the combustion engine is calculated as;

Efficiency = \frac{0utput \ work}{1nput \ work} \times 100\%\\\\Efficiency = \frac{356 \ \times \ 10^3}{946 \ \times \ 10^3} \times 100\%\\\\Efficiency = 37.63 \ \%

Therefore, the efficiency of this engine is 37.63 %.

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Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diamete
Sonbull [250]

Answer:

<em>work done in pumping the entire fuel is 466587 J</em>

<em></em>

Explanation:

weight of the gasoline per volume = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 2 m

height of the tractor tank above the top of the tank = 5 m

work done in pumping fuel to this height = ?

First, we find the volume of the fuel

since the tank is cylindrical,<em> we assume that the fuel within also takes the cylindrical shape.</em>

<em>Also, we assume that the fuel completely fills the tank.</em>

volume of a cylinder = \pi r^{2}l

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x 1.5^{2} x 2 = 14.139 m^3

we then find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

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8 0
2 years ago
Calculate the nuclear binding energy per nucleon for 136^Ba if its nuclear mass is 135.905 amu.
kompoz [17]

Answer:

1.312 x 10⁻¹² J/nucleon

Explanation:

mass of ¹³⁶Ba = 135.905 amu

¹³⁶Ba contain 56 proton and 80 neutron

mass of proton = 1.00728 amu

mass of neutron = 1.00867 amu

mass of ¹³⁶Ba = 56 x  1.00728 amu + 80 x 1.00867 amu

                      = 137.10128 amu

mass defect = 137.10128 - 135.905

                    = 1.19628 amu

mass defect = 1.19628 x 1.66 x 10⁻²⁷ Kg

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binding energy,

E = mass defect x c²

E = 1.9858 x 10⁻²⁷ x (3 x 10⁸)²

E = 17.87 x 10⁻¹¹ J/atom

now,

binding energy per nucleon =\dfrac{17.87\times 10^{-11}}{136}

                                              = 0.1312 x 10⁻¹¹ J/nucleon

                                              = 1.312 x 10⁻¹² J/nucleon

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