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2 years ago

What are common names assigned to instruction addresses in a PLC program called

Engineering

1 answer:

melomori [17]2 years ago

6 0

Answer:

thanks for the points.

Explanation:

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How to get on your screen on 2k20 in every mode

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The air in a room has a pressure of 1 atm, a dry-bulb temperature of 24C, and a wet-bulb temperature of 17C. Using the psychrome

TEA [102]

Answer:

(a) Relative Humidity = 48%,

Specific humidity = 0.0095

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3 years ago

A continuous random variable, X, whose probability density function is given by f(x) = ( λe−λx , if x ≥ 0 0, otherwise is said t

Ganezh [65]

Answer:

a) $F(x) = \lambda \int_0^{\infty} e^{-\lambda x} dx= -e^{-\lambda x} \Big|_0^{\infty} = 1- e^{-\lambda x} \$

b) $P(10 < X$

Explanation:

Previous concepts

The cumulative distribution function (CDF) F(x),"describes the probability that a random variableX with a given probability distribution will be found at a value less than or equal to x".

The exponential distribution is "the probability distribution of the time between events in a Poisson process (a process in which events occur continuously and independently at a constant average rate). It is a particular case of the gamma distribution".

Part a

Let X the random variable of interest. We know on this case that $X\sim Exp(\lambda)$

And we know the probability denisty function for x given by:

$f(x) = \lambda e^{-\lambda x} , x\geq 0$

In order to find the cdf we need to do the following integral:

$F(x) = \lambda \int_0^{\infty} e^{-\lambda x} dx= -e^{-\lambda x} \Big|_0^{\infty} = 1- e^{-\lambda x} \$

Part b

Assuming that $X \sim Exp(\lambda =0.1)$ , then the density function is given by:

$f(x) = 0.1 e^{-0.1 x} dx , x\geq 0$

And for this case we want this probability:

$P(10 < X$

And evaluating the integral we got:

$P(10 < X$

4 0

3 years ago

1. Design a circuit, utilizing set/reset coils where PB 1 starts Motor 1 and PB2 stops Motor 1. Pressing and releasing either pu

lianna [129]

Answer:

Circuit attached with explanation

Explanation:

Hi Dear,

A circuit is attached for your reference.

When you press "start" PB, the supply reaches the motor starter relay coil "M" that is also in parallel with the "start" PB which allows the motor to remain ON even when you release "start" PB as supply to relay coil is directly from supply "L" through "M".

To stop motor just press "stop" PB and the circuit breaks which de-energize the relay coil and the motor stops.

Hope this finds easy to you.

5 0

3 years ago