An ideal heat engine runs with a high temperature reservoir at 750K and a low temperature reservoir at 350K. When the engine is
1 answer:
Answer:
150 joules of energy is expelled to the low temperature reservoir each minute.
Explanation:
Lets assign the terms first.
where ...
= 750 K that is, the high temperature of the reservoir.
= 350 K, low temperature of the reservoir.
= 400 J,energy extract from the reservoir.
= 250 J, work done.
= energy expelled.
And
Formula to be used:
⇒
Re-arranging.
⇒
Now plugging the values of and
.
⇒
⇒ Joules
So energy expelled to the low temperature reservoir is of 150 joules.
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Answer:
Explanation:
We first identify the elements of this simple harmonic motion:
The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.
The angular frequency ω can be calculated with the formula:
Where k is the spring constant and m is the mass of the particle.
Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:
Finally, the equation of the motion of the system is:
or
Answer:
F = 63N
Explanation:
M= 1.5kg , t= 2s, r = (2t + 10)m and
Θ = (1.5t² - 6t).
magnitude of the resultant force acting on 1.5kg = ?
Force acting on the mass =
∑Fr =MAr
Fr = m(∇r² - rθ²) ..........equation (i)
∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)
The horizontal path is defined as
r = (2t + 10)
dr/dt = 2, d²r/dt² = 0
Angle Θ is defined by
θ = (1.5t² - 6t)
dθ/dt = 3t, d²θ/dt² = 3
at t = 2
r = (2t + 10) = (2*(2) +10) = 14
but dr/dt = 2m/s and d²r/dt² = 0m/s
θ = (1.5(2)² - 6(2) ) = -6rads
dθ/dt =3(2) - 6 = 0rads
d²θ/dt = 3rad/s²
substituting equation i into equation ii,
Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)
∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]
∑F = 1.5(14*3+0) = 63N
F = √(Fr² +FΘ²) = √(0² + 63²) = 63N
