Answer:
Wnet, in, = 133.33J
Explanation:
Provided that
Pump heat QH = 1000J
Hot temperature TH= 300K
Cold temperature TL= 260K
Given the heat pump is entirely reversible, the performance coefficient expression is formulated as follows:
According to the first law of thermodynamics,
COP(HP, rev) = 1/(1-TL/TH)
COP(HP, rev) = 1/(1-260/300)
COP(HP, rev) = 1/(1-0.867)
COP(HP, rev) = 1/0.133
COP(HP, rev) = 7.5
The power necessary to operate the heat pump is given by
Wnet, in = QH/COP(HP, rev)
Wnet, in = 1000/7.5
Wnet, in = 133.333J. QED
Thus, the 133.33J represents the initial work input during the heat transfer process.
<padditionally...><pbased on="" the="" first="" law="" rate="" at="" which="" heat="" is="" extracted="" from="" lower="" temperature="" reservoir="" calculated="" as="">
QL=QH-Wnet, in
QL=1000-133.333
QL=866.67J
</pbased></padditionally...>
The answer is 10pi. I believe this will be helpful.
Answer:
K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x
Explanation:
To calculate the kinetic energy variation, we can utilize the work-energy theorem.
W = ΔK
∫ F .dx = K - K₀
If the object starts from rest, then K₀ = 0.
So, ∫ F dx cos θ = K.
As the force and displacement directions align, the angle is zero, and hence the cosine is 1.
Now we can substitute and perform integration:
α ∫ x³ dx + β ∫ dx = K.
Thus, α x⁴ / 4 + β x = K.
Next, we evaluate from the limits F = 0 to F:
α (x⁴ / 4 - 0) + β (x - 0) = K.
Consequently, K = αX⁴ / 4 + β x.
This results in K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x.
To finalize the computation, we need to ascertain the displacement.
The force exerted on the car during the stop measures 6975 N.
Explanation: Given that the mass (m) is 930 kg, speed (s) at 56 km/h converts to 15 m/s, and the stopping time (t) is 2 s, we compute the force using F = m * a. Here, acceleration (a) can be obtained through a = s/t. The total force calculation confirms that F = 930 kg * (15 m/s) / 2 s results in 6975 N.
