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Korvikt [17]
3 years ago
11

All of the following are categories for clutch covers except

Engineering
1 answer:
Lostsunrise [7]3 years ago
7 0

Whichever one is not one of these.

Molded disc. This type of disc produces less shock when the clutch engages. ...

Metal disc. Metal material is durable and therefore offers longer life cycle. ...

Single-plate type. This refers to clutches that use a single clutch disc between the flywheel and pressure plate. ...

Dual-plate type.

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The minimum requirements for engineering documents are enumerated in
vladimir1956 [14]

Answer:

The answer will be Rule 61G15-23 F.A.C, relating to Seals.

Explanation:

According to the description given by: Florida administrative code&Florida administrative register the Minimum requirements for engineering documents are in the section 'Final 61G15-23'  from 11/3/2015. This document provides specifications of materials required for the safe operation of the system that is the result of engineering calculations, knowledge and experience.

8 0
3 years ago
A closed system consists of 0.3 kmol of octane occupying a volume of 5 m³. Determine (a) the weight of the system, in N, and (b)
Leni [432]

Answer:

a) m=336.18N

b) Vn=16.67m/kmol

Vm=0.1459m^3/kg

Explanation:

To calculate the mass of the octane(m):

Number of mole of octane (n) =0.3kmol(given)

Molarmass of octane (M) =114.23kg/kmol

m=n*M

m=(0.3kmol)*(114.23kg/kmol)

m=34.269kg

To calculate for the weight of octane(W):

W=g*m

W=(9.81m/s^2)*(34.269kg)

W=336.18N

b) For specific volumes of Vn and Vm:

Given volume of octane (V) =5m^3

Vm=V/m

Vm=5m^3/34.269kg

Vm=0.1459m^3/kg

And Vn will be :

Vn=V/m=5m^3/0.3kmol

Vn=16.67m/Kmol

Therefore, the answers are:

a) m=336.18N

b) Vn=16.67m/kmol

Vm=0.1459m^3/kg

7 0
3 years ago
Consider a half-wave rectifier circuit with a triangular-wave input of LaTeX: 6V6 V peak-to-peak amplitude and zero average valu
Varvara68 [4.7K]

The question is not complete. We are supposed to find the average value of v_o.

Answer:

v_o,avg = 0.441V

Explanation:

Let t1 and t2 be the start and stop times of the output waveforms. Thus, from the diagram i attached, using similar triangles, we have;

3/(T/4) = 0.7/t1

So, 12/T = 0.7/t1

So, t1 = 0.7T/12

t1 = 0.0583 T

Also, from symmetry of triangles,

t2 = T/2 - t1

So, t2 = T/2 - 0.0583 T

t2 = 0.4417T

Average of voltage output is;

v_o,avg = (1/T) x Area under small triangle

v_o,avg = (1/T) x (3 - 0.7) x (T/4 - t1)

v_o,avg = (1/T) x (2.3) x (T/4 - 0.0583 T)

v_o,avg = (1/T) x 2.3 x 0.1917T

T will cancel out to give;

v_o,avg = 0.441V

8 0
2 years ago
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