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3 years ago

What are the important factors needed to be considered while selecting a brake or clutch?

Engineering

1 answer:

AysviL [449]3 years ago

4 0

Answer:

The correct answer is: the following factors are needed to properly consider while selecting a brake or clutch:

-Engagement

-Friction

-Electromagnetic

-Mechanical

-Actuation

-Electric

-Fluid power

-Self-actuation

-Key concepts

-Application notes

-Selection criteria

Explanation:

Clutches and brakes are important devices in many rotating drive systems, it is very important to guarantee the security and the proper function of them accomplishing a high quality parameters in those factors.

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You are driving on a roadway with multiple lanes of travel in the same direction, and are approaching an emergency vehicle parke

solmaris [256]

Answer: a. Leave the lane closest to the emergency as soon as it is safe to do so, or slow down to a speed of 20 MPH below the posted speed limit.

Explanation:

Giving a way to the law enforcement vehicle and a medical emergency vehicle is necessary. If one approaches an emergency vehicle parked along the roadway one should change the lane as the vehicle may not move and the driver may also waste his or her time also one should also slow down his or her speed while approaching the vehicle as most of the emergency vehicle are in rush to reach the hospital so the driver should maintain some distance with the medical emergency vehicle.

7 0

3 years ago

Why is it important to keep hand tools free of oil and grease?

Bezzdna [24]

Answer:

keeps it from rusting

Explanation:

5 0

3 years ago

How does the "E" in STEM work with the other letters

KIM [24]

Answer:

if you are speaking of the acronym then Engineering uses science and mathematics to solve everyday problems in society

4 0

3 years ago

Which of the following best distinguishes between superficial design improvements and functional

Contact [7]

Answer:

Superficial design improvements are typically only trivial changes to a design, while functional design improvements can change the way a product or process is used to significantly enhance performance.

Explanation:

As a PC board designer, I would sometimes spend a certain amount of time making traces have shorter routes, or fewer layer changes or bends. (I wanted to make the layout "pretty.") In some cases, these changes are superficial, affecting the appearance only. In some cases, they are functional, reducing crosstalk or emissions or susceptibility to interference.

I deal with a web site that seems to be changing all the time (Brainly). In many cases, the same information is rearranged on the page—a superficial change. In other cases, the information being displayed changes, or the way that certain information is accessed changes. These are functional changes. (Sometimes, they "enhance performance," and sometimes they don't, IMO.)

In short ...

<em>Superficial design improvements are typically only trivial changes to a design, while functional design improvements can change the way a product or process is used to significantly enhance performance.</em>

8 0

2 years ago

120 litres of water is discharge from container in 25 seconds. Find the rate of discharge in cumecs.if the discharge took place

notsponge [240]

<h2>Answer:</h2>

Rate of discharge in cumecs: <u>0.0048m³/s</u>.

Velocity flow: <u>24m/s</u>.

<h2>Explanation:</h2>

<h3>1. Find the rate of discharge in cumecs.</h3>

a. Convert from litres to m³.
120L*1000= 120000mL

120000mL=120000cm³

120000cm³/100³=0.12 m³.

b. Rate of discharge.

<em>If 0.12 m³ where discharged in 25 seconds, the rate of discharge is:</em>

0.12m³/25s = 0.0048m³/s.

<em />

<h3>2. Find the velocity flow.</h3>

Let's refer to the fluid mechanics equation that relates volume flow, area and velocity. This is the formula:

$\frac{dV}{dt}=Av$ ; where the expression $\frac{dV}{dt}$ is the volume flow rate (in m³/s); A is the cross-sectional area of the pipe (in m²), and v is the velocity flow (in m/s).

a. Solve the equation for v.

$\frac{dV}{dt}=Av\\ \\(\frac{dV}{dt})/A=v\\ \\v=(\frac{dV}{dt})/A$

b. Calculate the cross-sectional area of the pipe.

<em>The cross-sectional area of the pipe is a circle. Hence, the formula of this area is:</em>

$A=\pi r^{2}$

<em>We'll have to convert the diameter to meters, because the formula for flow velocity needs the area in m². Let's go ahead and do that.</em>

<em /> $50mm/1000=0.05m$ .

<em>We were given the diameter, and the formula uses the radius, but the radius is just half of the diameter, therefore, we can substitute in toe formula like this:</em>

$A=\pi (\frac{0.05}{2} )^{2}=0.0020m^{2}$

c. Substitute in the new expression for velocity flow and calculate.

$v=(\frac{dV}{dt})/A\\ \\v=(\frac{0.048m^{3} }{1s})/(0.0020m^{2} )\\\\ v= 24m/s$

8 0

1 year ago