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2 years ago

L2

1 answer:

4 0

There are different kinds of statements. The statement that is true regarding the process-control relay ladder diagram shown in the figure (Image attached) is that;

Only the temperature switch has to close for the motor to start.

Note that the pressure switch and the push button often occurs in stages or series and it is known to be parallel with the temperature switch.

The Relay Ladder Logic is known to be a primary programming language that is often used for programmable logic controllers (PLCs).

The ladder diagram often makes use of the contacts to show the switches, or any input. Note that the coil symbol often depicts the output. There is a line that is depicting an input/output and it is called a rung.

Pressure switches are said to be key parts that are used for managing or controlling the activation and deactivation of pumps in fluid systems.

Learn more about Ladder from

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Which of these rotational quantities is analogous to mass in a linear system?

Rom4ik [11]

Answer:

Explanation:

i think it is rotational inertia

because analogue of mass in rotational motion is moment of inertia. It plays the same role as mass plays in transnational motion.

hope it's right & helps !!!!!!!!!

5 0

2 years ago

The potential difference between two equipotential lines 5.0 mm apart has a value of 1.2 V, calculate the electric field value.

jonny [76]

Answer:

The electric field value is 240 N/C

Explanation:

Given that,

Distance = 5.0 mm

Potential difference = 1.2 V

We need to calculate the electric field value

Using formula of potential difference

$\Delta V=Ed$

$E=\dfrac{\Delta V}{d}$

Where, E = electric field

V = potential difference

d = distance

Put the value into the formula

$E=\dfrac{1.2}{5.0\times10^{-3}}$

$E= 240\ N/C$

Hence, The electric field value is 240 N/C

6 0

3 years ago

When the mass is removed, the length of the cable is found to be l0 = 4.66 m. After the mass is added, the length is remeasured

zaharov [31]

Answer:

$\gamma=6.07*10^5\frac{N}{m^2}$

Explanation:

For a linear elastic material Young's modulus is a constant that is given by:

$\gamma=\frac{F/A}{\Delta L/L_0}$

Here, F is the force exerted on an object under tensio, A is the area of the cross-section perpendicular to the applied force, $\Delta L$ is the amount by which the length of the object changes and $L_0$ is the original length of the object. In this case the force is the weight of the mass:

$F=mg\\F=55kg(9.8\frac{m}{s^2})\\F=539N$

Replacing the given values in Young's modulus formula:

$\gamma=\frac{F/\pi r^2}{(L_1-L_0)/L_0}\\\gamma=\frac{539N/\pi(0.045m)^2}{(5.31m-4.66m)/4.66m}\\\gamma=6.07*10^5\frac{N}{m^2}$

6 0

2 years ago

A(n) 93 kg clock initially at rest on a horizontal floor requires a(n) 610 N horizontal force to set it in motion. After the clo

denpristay [2]

Answer:0.669

Explanation:

Given

mass of clock 93 kg

Initial force required to move it 610 N

After clock sets in motion it requires a force of 514 N to keep moving it with a constant velocity

Initially static friction is acting which is more than kinetic friction

thus 613 force is required to overcome static friction

$\mu _smg=610$

$\mu _s\times 93\times 9.8=610$

$\mu _s=0.669$

5 0

3 years ago

At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is

Helen [10]

Answer:

The thrown rock will strike the ground $2.42s$ earlier than the dropped rock.

Explanation:

Known Data

$y_{i}=300m$

$y_{f}=0m$
$v_{iD}=0m/s$
$v_{iT}=-29m/s$ , it is negative as is directed downward

Time of the dropped Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}$ , then clearing for $t_{D}$ .

$t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s$

Time of the Thrown Rock

We can use $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find the total time of fall, so $0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}$ , then, $0=-4.9t_{T}^{2}-29t_{T}+300$ , as it is a second-grade polynomial, we find that its positive root is $t_{T}=5.4s$

Finally, we can find how much earlier does the thrown rock strike the ground, so $t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s$

6 0

3 years ago