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Gelneren [198K]
2 years ago
13

Compare and contrast mechanical properties of plastics, metals and ceramics.

Engineering
1 answer:
IgorLugansk [536]2 years ago
6 0

Answer:

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Explanation:

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electrial systems

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Relay contacts that are defined as being normally open (n.o.) have contacts that are:_____.
Reptile [31]

Relay contacts that are defined as being normally open (n.o.) have contacts that are open only if  the relay coil is known to have de-energized.

<h3>What is meant by normally open contacts?</h3>

Normally open (NO) are  known to be open if there is no measure of current that is flowing through a given coil but it often close as soon as the coil is said to be energized.

Note that  Normally closed (NO) contacts are said to be closed only if the coil is said to be de-energized and open only if the coil is said to carry current or is known to have energized.

The role of relay contact is wide. The Relays are tools that are often used in the work of  switching of control circuits and it is one that  a person cannot used for power switching that has relatively bigger ampacity.

Therefore, Relay contacts that are defined as being normally open (n.o.) have contacts that are open only if  the relay coil is known to have de-energized.

Learn more about Relay contacts from

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8 0
1 year ago
Two vertical, parallel clean glass plates are spaced a distance of 2mm apart. if the plates are placed in water, how high will t
Ulleksa [173]

Answer with Explanation:

The capillary rise in 2 parallel plates immersed in a liquid is given by the formula

h=\frac{2\sigma cos(\alpha )}{\rho gd}

where

\sigma is the surface tension of the liquid

\alpha is the contact angle of the liquid

\rho is density of liquid

'g' is acceleratioj due to gravity

'd' is seperation between thje plates

Part a) When the liquid is water:

For water and glass we have

\sigma =7.28\times 10^{-2}N/m

\alpha =0

\rho _{w}=1000kg/m^3

Applying the values we get

h=\frac{2\times 7.28\times 10^{-2}cos(0)}{1000\times 9.81\times 2\times 10^{-3}}=7.39mm

Part b) When the liquid is mercury:

For mercury and glass we have

\sigma =485.5\times 10^{-3}N/m

\alpha =138^o

\rho _{w}=13.6\times 10^{3}kg/m^3

Applying the values we get

h=\frac{2\times 485.5\times 10^{-3}cos(138)}{13.6\times 1000\times 9.81\times 2\times 10^{-3}}=-2.704mm

The negative sign indicates that there is depression in mercury in the tube.

4 0
3 years ago
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