4 water has a low heat capacity and a high vaporization temperate and coasts have low temps
Answer:
the work is done by the gas on the environment -is W= - 3534.94 J (since the initial pressure is lower than the atmospheric pressure , it needs external work to expand)
Explanation:
assuming ideal gas behaviour of the gas , the equation for ideal gas is
P*V=n*R*T
where
P = absolute pressure
V= volume
T= absolute temperature
n= number of moles of gas
R= ideal gas constant = 8.314 J/mol K
P=n*R*T/V
the work that is done by the gas is calculated through
W=∫pdV= ∫ (n*R*T/V) dV
for an isothermal process T=constant and since the piston is closed vessel also n=constant during the process then denoting 1 and 2 for initial and final state respectively:
W=∫pdV= ∫ (n*R*T/V) dV = n*R*T ∫(1/V) dV = n*R*T * ln (V₂/V₁)
since
P₁=n*R*T/V₁
P₂=n*R*T/V₂
dividing both equations
V₂/V₁ = P₁/P₂
W= n*R*T * ln (V₂/V₁) = n*R*T * ln (P₁/P₂ )
replacing values
P₁=n*R*T/V₁ = 2 moles* 8.314 J/mol K* 300K / 0.1 m3= 49884 Pa
since P₂ = 1 atm = 101325 Pa
W= n*R*T * ln (P₁/P₂ ) = 2 mol * 8.314 J/mol K * 300K * (49884 Pa/101325 Pa) = -3534.94 J
Answer:
An asteroid impact could affect the tilt of the Earth due to the force it applies onto the planet. This would change Earth's seasons due to the fact that Earth's tilt causes seasons.
Answer:
a) -1.25 rev/s² and 23.3 rev
b) 2.67s
Explanation:
a) ω
= (500 rev/min)(1min/ 60s) => 8.333 rev/s
ω
= (200 rev/min)(1min/ 60s) => 3.333rev/s
time 't'= 4 s
angular acceleration 'α'=?
constant angular acceleration equation is given by,
ω= ω + αt
α= (ω - ω )/t => (3.333-8.333)/4
α= -1.25 rev/s²
θ-θ
= ω t + 1/2αt²
=(8.333)(4) + 1/2 (-1.25)(4)²
=23.3 rev
b) ω=0 (comes to rest)
ω = 3.333 rev/s
α= -1.25 rev/s²
ω= ω + αt
t= (ω - ω)/α => (0- 3.333)/-1.25
t= 2.67s
Answer:
A) 
B) F = 1632.65 N
Explanation:
Given details
outside air speed is given as 
since inside air is atmospheric , 
a) By using bernoulli equation between outside and inside of flight
![\Delta P = \frac{1}{2} \rho[ v_2^2 -v_1^2]](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%5Crho%5B%20v_2%5E2%20-v_1%5E2%5D)
![\Delta P = \frac{1}{2} 1.29 [ 150^2 - 0^2]](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20%5Cfrac%7B1%7D%7B2%7D%201.29%20%5B%20150%5E2%20-%200%5E2%5D)
b) force exerted on window
Area of window 
We know that force is given as
F = 1632.65 N
