Answer:
A three valve manifold serves to protect the capsule from being exposed to pressures above the acceptable range and also enables transmitter isolation from the loop of the process
In closed tank level measurement installations the pressure of the gas if presence of gas on top of the liquid will need to be taken account by the use of a reference connection from the level transmitter to the tank top. The valve used in pressure equalization is kept close when the process system is operating normally.
In the event that the transmitter is removed or put into service, it is necessary to ensure that the gas phase which is at the higher pressure is does not over-range the Differential Pressure DP capsule
Explanation:
Answer:
-0.55m/s
Explanation:
Given that: For the boy
Weight = 745N
Velocity = +0.35 m/s
Mass of the boy = ?
g = 9.81m/s^2
W = mg
745 = m×9.81
m = 75.94kg
For the girl
Given that:
Weight = 477 N
g = 9.81m/s^2
m = ?
W = mg
477 = m×9.81m/s^2
m = 48.62kg
To solve for the v of the girl, the two has to add up
48.62kg×v + 75.94kg×+0.35 m/s = 0
48.62v + 26.579 = 0
48.62v = - 26.579
v = -26.579/48.62
v = -0.5466
v = -0.55m/s
Hence, the velocity of the girl is -0.55m/s.
The negative sign is as a result of the two of them moving is opposite direction.
Water that is "hard" <span>contains a certain amount of dissolved minerals (like calcium and magnesium).
</span>Water that is "soft" <span>is a certain type of treated </span>water. It contains <span>only one type of ion, which is sodium. An example of this is rainwater because it is naturally </span>soft water.
I would say 1000c as my answer
<h2>
The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Time, t = 3 s
Substituting
s = ut + 0.5 at²
s = 0 x 3 + 0.5 x 9.81 x 3²
s = 44.145 m
The seagull's approximate height above the ground at the time the clam was dropped is 4 m
