Answer:
The pressure after passing the valve is 23,8 [Kpa] ( 0,234 atm) and the pressure drop is about 1,53 [Kpa]
Explanation:
We need to use the formula of bernoulli, in the attached image we can see the fluid throw the pipe, we also can calculate the velocity inside the pipe using the flow rate and the cross sectional area.
For this case, we don't use the elevation difference and therefore those terms can be cancelled.
When the area has reduced the velocity of the fluid is increased but there is a drop pressure through the valve.
The elements is group 18 on the periodic table are called "Nobel Gases".
Under the assumption that the tires do not change in volume, apply Gay-Lussac's law:
P/T = const.
P = pressure, T = temperature, the quotient of P/T must stay constant.
Initial P and T values:
P = 210kPa + 101.325kPa
P = 311.325kPa (add 101.325 to change gauge pressure to absolute pressure)
T = 25°C = 298.15K
Final P and T values:
P = ?, T = 0°C = 273.15K
Set the initial and final P/T values equal to each other and solve for the final P:
311.325/298.15 = P/273.15
P = 285.220kPa
Subtract 101.325kPa to find the final gauge pressure:
285.220kPa - 101.325kPa = 183.895271kPa
The final gauge pressure is 184kPa or 26.7psi.
Answer:
The magnification is a function of the lenses in the objective and the eyepiece, so the magnification of the two must be multiplied to obtain the total magnification possible. So, for example, if the objective lens was 4X and the eye piece lens was 10X, the total magnification would be 40. (4 x 10 = 40)
Explanation:
Answer:
v = 83.1 % of speed of light
Explanation:
given,
T_e is the earth time = 2.7 s
T_s is the ship time = 1.5 s
we know,
where c is the speed of light
v is the speed of the rock star moving
squaring both side
v = 0.831 c
v = 83.1 % of speed of light
