Answer:
Depth of the pool, h = 4.004 cm
Explanation:
Pressure at the bottom, P = 39240 N/m²
The density of water, d = 1000 kg/m³
The pressure at the bottom is given by :
P = dgh
We need to find the depth of pool. Let h is the depth of the pool. So,


h = 4.004 m
So, the pool is 4.004 meters pool. Hence, this is the required solution.
Answer:
check the point where the output device's field wiring is connected to the output rack. ( C )
Explanation:
Given that an indicator light on the output module is indicating that a signal has been sent to the output point where the failed single output device is connected , the best line of action to properly indicate the output device that has filed in the PLC system, is to check the point where the output device's field wiring is connected to the output rack in the PLC system.
A PLC system is a computer control system ( especially Industrial ) that helps in the monitoring the states of input and output devices in an industry.
32f. That's because the force is directly proportional to the product of the masses and inversely proportional to the square of the distance. So you get 2•(1/1/4)^2=2•16=32
The volume corresponds to the measure of the space occupied by a body. From the given dimensions we can intuit that we are looking to find the Volume of an Cuboid, that is, an orthogonal rectangular prism, whose faces form straight dihedral angles.
Mathematically the volume of this body is given as

Where,
L = Length
W = Width
H = High


Note: The value given for the height was in centimeters, so it was transformed to meters.
The force on a charged particle on a magnetic field is given by:

where B is the magnitude of the field, q is the charge of the particle, v is its velocity and θ is the angle between the field and the velocity of the particle.
In this case we have that:
• The magnitude of the field is 0.09 T.
,
• The charge of the particle is 3.2 x 10 –19C .
,
• The velocity is 3.0 × 10 4 m/s
,
• The angle between the field and the velocity is 90°, since they are perpendicular.
Plugging these we have:

Therefore, the force on the charge is: