A bus goes from rest to 20 m/s. The bus accelerates at 2 m/s^2. How far does the bus go?
1 answer:
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Answer:
The change in internal energy of the air within the piston is 490 J.
Explanation:
In thermodynamics the internal energy (ΔU) of a system is the total energy contained in the system and can be considered as the sum of the energy in the form of heat (q, given to or released by the system) and in the form of work (w, make by the system to the environment or from the environment to the system). When the system absorbs heat from the enviroment this value is positive, while when the system realeased heat to the enviroment, the value is negative. In the case of work, when the enviroment make work on the system, the value is positive and, on the contrary, when the system makes work against the environment the value is negative.
ΔU = q + w
w is defined as the negative product between the external pressure (P) and the change in volume (ΔV). [w = - P·ΔV]
⇒ ΔU = q - P·ΔV
In this problem, the system is composed by the cylinder + piston + contained air (Attached)
1) q= 565 J (a positive value because the enviroment delivered heat to the cylinder)
2) P = 1 atm.
3) ΔV = 0.86 atm.L - 0.12 atm.L = 0.74 atm.L.
⇒ ΔU = q + w ⇒ ΔU = 565 J - (1 atm)·(0.74 L) ⇒ ΔU = 565 J - 0.74 atm.L
We have to express the work value in the same units of heat, it means Joules. As one joule is equal to 0.00987 atm.L
⇒ (0.74 atm.L) x ( 1 J/0.00987 atm.L) = 74.97 J.
⇒ ΔU = q + w ⇒ ΔU = 565 J - 74.97 J ⇒ ΔU = 490 J.
Summarizing, the change in internal energy of the air within the piston is 490 J.
Answer:
a) 0.00996 m
b) 109090909 Pa
Explanation:
Unit conversions:
1.2 mm = 0.0012 m
8.5 kN = 8500 N
If the 2.2m rod cannot stretch more than 0.0012 m, its maximum strain is
With elastic modulus being E = 200 GPa, then its maximum stress must be
Knowing the tension force being F = 8500 N, we can calculate the appropriate cross section area
And its corresponding diameter is