Answer:
74 Ω
Explanation:
Data provided in the question:
Maximum value of the current that can be provided = 500 mA
= 500 × 10⁻³ A
Applied voltage set for the system = 37 V
Now,
The smallest resistance that can be measured
= [ Applied voltage ] ÷ [ Maximum current that can be applied ]
= 37 ÷ ( 500 × 10⁻³ )
= 74 Ω
Answer:
Yes, the flow is turbulent.
Explanation:
Reynolds number gives the nature of flow. If he Reynolds number is less than 2000 then the flow is laminar else turbulent.
Given:
Diameter of pipe is 10mm.
Velocity of the pipe is 1m/s.
Temperature of water is 200°C.
The kinematic viscosity at temperature 200°C is
m2/s.
Calculation:
Step1
Expression for Reynolds number is given as follows:

Here, v is velocity,
is kinematic viscosity, d is diameter and Re is Reynolds number.
Substitute the values in the above equation as follows:


Re=64226.07579
Thus, the Reynolds number is 64226.07579. This is greater than 2000.
Hence, the given flow is turbulent flow.
Answer: Option D is not true of hydraulic valves. A hydraulic valve is a device that can change the opening degree of liquid flow path
Explanation:
The pilot check valve allows flow of liquid in one direction and blocks flow in the opposite direction
To resolve this problem we have,

is unknown.
With these dates we can calculate the Flexural strenght of the specimen,

After that, we can calculate the flexural strenght for the square cross section using the previously value.

Answer:
The time taken will be "1 hour 51 min". The further explanation is given below.
Explanation:
The given values are:
Number of required layers:
= 
= 
Diameter (d):
= 1.25 mm
Velocity (v):
= 40 mm/s
Now,
The area of one layer will be:
= 
= 
The area covered every \second will be:
= 
= 
= 
The time required to deposit one layer will be:
= 
= 
The time required for one layer will be:
= 
∴ Total times required for one layer will be:
= 
= 
So,
Number of layers = 152
Therefore,
Total time will be:
= 
= 
= 