Consider a short time span just before and after the spark plug in a gasoline engine ignites the fuel-air mixture and releases 1
2 answers:
Answer:
The temperature after ignition is 7859.565 K
Explanation:
Here we have
Energy released = 1970 J
Volume of gas = 47.9 cm³ = 0.0000479 m³
Pressure of gas = 1.18 × 10⁶ Pa
Temperature of he gas = 325 K
From P·V = n·R·T
Therefore, n = PV/(RT) = (0.0000479 × 1.18 × 10⁶)/(8.3145× 325) = 2.09 × 10⁻² moles
For an ideal mono-atomic gas, the mono-atomic gas, the molar heat capacity at constant volume is given as
= 3/2 R ≈ 12.5 J K−1 mol⁻¹
Therefore we have
Heat released, ΔH = 1970 J = n × ×(T₂ - T₁) = 2.09 × 10⁻² × 12.5×(T₂ -325)
T₂ = (1970 J)/(2.09 × 10⁻² × 12.5) + 325
= 7534.565 +325 = 7859.565 K
The temperature after ignition = 7859.565 K.
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Answer:
The <em>net gravitational force it exerts</em> is
Explanation:
Newton's Law of Gravitation can be written as
where <em>G is the Gravitational Constant, m1 and m2 are the masses of two objects, and r is the distance between them</em>. In this case, the spheres are loacted in straight line, so instead of a vector r, we have a distance x in meters. The distances and masses are given in the problem, and the smaller sphere is between the other two spheres. This means <u>the sphere 1 is in the middle, the sphere 2 is on the left of 1, and the sphere 3 is on the right of 1</u>, so
is the force that 2 feels because of 1, and
is the force that 3 feels because of 1.
<em>If we replace the data in those previous equations</em>, we have that
Finally, adding both results, the net force the sphere 1 exerts is