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A single output device has failed while the remainder of the PLC system is functioning normally The indicator light on the outpu

Dmitry_Shevchenko [17]

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.

8 0

4 years ago

A particle is moved along the x-axis by a force that measures 10/(1+x)^2 pounds at a point x feet from the origin. Find the work

jarptica [38.1K]

Answer:

9 ft*lb

Explanation:

super simple but you just have to understand that the integral is going with the curve

work = integral a to b of f(x)dx = integral 0 to 9 of 10/(1+x)^2dx = 9ft*lb

6 0

3 years ago

If the block is subjected to the force of F = 500 N, determine its velocity at s = 0.5 m. When s = 0, the block is at rest and t

STatiana [176]

Answer:

The velocity is $4.6 m/s^2$

Explanation:

Given:

Force = 500N

Distance s= 0

To find :

Its velocity at s = 0.5 m

Solution:

$\sum F_{x}=m a$

$F\left(\frac{4}{5}\right)-F_{S}=13 a$

$500\left(\frac{4}{5}\right)-\left(k_{s}\right)=13 a$

$400-(500 s)=13 a$

$a = \frac{400 -(500s)}{13}$

$a = (30.77 -38.46s) m/s^2$

Using the relation,

$a=\frac{d v}{d t}=\frac{d v}{d s} \times \frac{d s}{d t}$

$a=v \frac{d v}{d s}$

$v d v=a d s$

Now integrating on both sides

$\int_{0}^{v} v d v=\int_{0}^{0.5} a d s$

$\int_{0}^{v} v d v=\int_{0}^{0.5}(30.77-38.46 s) d s$

$\left[\frac{v^{2}}{2}\right]_{0}^{2}=\left[\left(30.77 s-19.23 s^{2}\right)\right]_{0}^{0.5}$

$\left[\frac{v^{2}}{2}\right]=\left[\left(30.77(0.5)-19.23(0.5)^{2}\right)\right]$

$\left[\frac{v^{2}}{2}\right]=[15.385-4.807]$

$\left[\frac{v^{2}}{2}\right]=10.578$

$v^{2}=10.578 \times 2$

$v^{2}=21.15$

$v = \sqrt{21.15}$

$v = 4.6 m/s^2$

8 0

3 years ago

You are riding your bike to the mall. You travel the first mile in 10 minutes. The last mile takes you 15 minutes. This is an ex

Vikki [24]

This is an example of slowing down because you encountered heavy traffic or just got tired. Also, the numbers show acceleration.

6 0

4 years ago

A 500-watt vacuum cleaner is plugged into a 120-volt outlet and used for 30 minutes. How much current runs through the vacuum?

Ymorist [56]

   Power = (voltage) x (current)

   500 W = (120 v) x (current)

   Current = (500 w) / (120 v) = 4-1/6 amperes .

Power and current are both rates.
Power is the rate of using energy ... joules per second.
Current is the rate of flowing charge ... coulombs per second.
Neither quantity depends on how long you keep it up.
They're both going on continuously while the vacuum is running.

4 0

4 years ago

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