Answer:
a)W= - 720 J
b)ΔU= 330 J
Explanation:
Given that
P = 0.8 atm
We know that 1 atm = 100 KPa
P = 80 KPa
V₁ = 12 L = 0.012 m³ ( 1000 L = 1 m³)
V₂ = 3 L = 0.003 m³
Q= - 390 J ( heat is leaving from the system )
We know that work done by gas given as
W = P (V₂ -V₁ )
W= 80 x ( 0.003 - 0.012 ) KJ
W= - 0.72 KJ
W= - 720 J ( Negative sign indicates work done on the gas)
From first law of thermodynamics
Q = W + ΔU
ΔU=Change in the internal energy
Now by putting the values
- 390 = - 720 + ΔU
ΔU= 720 - 390 J
ΔU= 330 J
Answer:
Take-off velocity = v = 81.39[m/s]
Explanation:
We can calculate the takeoff speed easily, using the following kinematic equation.
where:
a = acceleration = 4[m/s^2]
x = distance = 750[m]
vi = initial velocity = 25 [m/s]
vf = final velocity
![v_{f}=\sqrt{(25)^{2}+(2*4*750) } \\v_{f}=81.39[m/s]](https://tex.z-dn.net/?f=v_%7Bf%7D%3D%5Csqrt%7B%2825%29%5E%7B2%7D%2B%282%2A4%2A750%29%20%7D%20%5C%5Cv_%7Bf%7D%3D81.39%5Bm%2Fs%5D)
A. a<span> = 1.3 m/s^2</span><span>; </span>FN<span> = 63.1 N</span>
