Since the temperature of the gas remains constant in the process, we can use Boyle's law, which states that for a gas transformation at constant temperature, the product between the gas pressure and its volume is constant:
which can also be rewritten as
(1)
where the labels 1 and 2 mark the initial and final conditions of the gas.
In our problem,
,
and
, so the final pressure of the gas can be found by re-arranging eq.(1):
Therefore the correct answer is
<span>1. 0.75 atm</span>
Answer:
The speed of sound, in m/s, through air at this temperature is 343.5 m/s
Explanation:
Given;
distance traveled by sound, d = 1,687.5 meters
time taken for the sound to travel, t = 5 seconds
air temperature, θ = 10°C
Speed of sound = distance traveled by sound / time taken for the sound to travel
Speed of sound = d / t
= 1687.5 m / 5 s
= 337.5 m/s
Speed of sound at the given temperature is calculated as;
c = 337.5 + 0.6θ
c = 337.5 + 0.6 x 10
c = 337.5 + 6
c = 343.5 m/s
Therefore, the speed of sound, in m/s, through air at this temperature is 343.5 m/s
9 is D I believe, I don't know about 10
Answer:
A) At point 1, local acceleration = 0.5 m/s²
At point 2, local acceleration = 1.0 m/s²
B) Average Eulerian convective acceleration over the two points in the cross section shown = 0.5 m/s²
This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.
Explanation:
Local acceleration at those points is the instantaneous acceleration at those points and it is given as
a = dv/dt
At point 1, v₁ = 0.5 t
a₁ =dv₁/dt = 0.5 m/s²
At point 2, v₂ = 1.0 t
a₂ = dv₂/dt = 1.0 m/s²
b) Average Eulerian convective acceleration over the two points in the cross section shown = (change of velocity between the two points)/time
Change of velocity between the two points = v₂ - v₁ = 1.0t - 0.5t = 0.5 t
Time = t
Average acceleration = 0.5t/t = 0.5 m/s²
This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.
Answer:
It does both. Once they get close enough the air does start to get charged, but then they eventually discharge when they touch.
Explanation:
