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stiks02 [169]
3 years ago
8

Which statement describes a third class lever?

Engineering
1 answer:
Sav [38]3 years ago
8 0

Answer:

The input force is between the fulcrum and the output force

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A batch of 1000 is split into 10 smaller batches of equal size 100. The processing time of each unit is 2
Vika [28.1K]

The lead time of the actual batch will be in

  • 2950 in minutes

<h3>What is Processing Time?</h3>

This refers to the amount of time which is taken for a processor to run a procedure and return a result.

We can see that a batch of 1000 is split so that they each have 10 smaller batches which has an equal size of 100 each, then if the processing time is 2 mins per machine and the set up time is 30 mins.

Hence, when this batch is processed over a serial line of 5 machines, then the lead time of the actual batch would be 2950 in minutes

Read more about processing time here:

brainly.com/question/18444145

4 0
2 years ago
A 500-km, 500-kV, 60-Hz, uncompensated three-phase line has a positivesequence series impedance. z = 5 0.03 1 + j 0.35 V/km and
Anni [7]

Answer:

A) 282.34 - j 12.08 Ω

B) 0.0266 + j 0.621 / unit

C)

A = 0.812 < 1.09° per unit

B =  164.6 < 85.42°Ω  

C =  2.061 * 10^-3 < 90.32° s

D =  0.812 < 1.09° per unit

Explanation:

Given data :

Z ( impedance ) = 0.03 i  + j 0.35 Ω/km

positive sequence shunt admittance ( Y ) = j4.4*10^-6 S/km

A) calculate Zc

Zc = \sqrt{\frac{z}{y} }  =  \sqrt{\frac{0.03 i  + j 0.35}{j4.4*10^-6 } }    

    = \sqrt{79837.128< 4.899^o}   =  282.6 < -2.45°

hence Zc = 282.34 - j 12.08 Ω

B) Calculate  gl

gl = \sqrt{zy} * d  

 d = 500

 z = 0.03 i  + j 0.35

 y = j4.4*10^-6 S/km

gl =  \sqrt{0.03 i  + j 0.35*  j4.4*10^-6}  * 500

   = \sqrt{1.5456*10^{-6} < 175.1^0} * 500

   = 0.622 < 87.55 °

gl = 0.0266 + j 0.621 / unit

C) exact ABCD parameters for this line

A = cos h (gl) . per unit  =  0.812 < 1.09° per unit ( as calculated )

B = Zc sin h (gl) Ω  = 164.6 < 85.42°Ω  ( as calculated )

C = 1/Zc  sin h (gl) s  =  2.061 * 10^-3 < 90.32° s ( as calculated )

D = cos h (gl) . per unit = 0.812 < 1.09° per unit ( as calculated )

where :  cos h (gl)  = \frac{e^{gl} + e^{-gl}  }{2}

             sin h (gl) = \frac{e^{gl}-e^{-gl}  }{2}

     

7 0
3 years ago
When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the str
Murljashka [212]

The question is incomplete. The complete question is :

The solid rod shown is fixed to a wall, and a torque T = 85N?m is applied to the end of the rod. The diameter of the rod is 46mm .

When the rod is circular, radial lines remain straight and sections perpendicular to the axis do not warp. In this case, the strains vary linearly along radial lines. Within the proportional limit, the stress also varies linearly along radial lines. If point A is located 12 mm from the center of the rod, what is the magnitude of the shear stress at that point?

Solution :

Given data :

Diameter of the rod : 46 mm

Torque, T = 85 Nm

The polar moment of inertia of the shaft is given by :

$J=\frac{\pi}{32}d^4$

$J=\frac{\pi}{32}\times (46)^4$

J = 207.6 mm^4

So the shear stress at point  A is :

$\tau_A =\frac{Tc_A}{J}$

$\tau_A =\frac{85 \times 10^3\times 12 }{207.6}$

$\tau_A = 4913.29 \ MPa$

Therefore, the magnitude of the shear stress at point A is 4913.29 MPa.

3 0
3 years ago
Why is it important to cut all the way through an electrical wire on the first try?
lbvjy [14]

Answer:

I always thought it was so that the older wire could not have a problem and have another electrician must come back and fix it.

Explanation:

6 0
3 years ago
A 100 ft long steel wire has a cross-sectional area of 0.0144 in.2. When a force of 270 lb is applied to the wire, its length in
blondinia [14]

Answer:

(a) The stress on the steel wire is 19,000 Psi

(b) The strain on the steel wire is 0.00063

(c) The modulus of elasticity of the steel is 30,000,000 Psi

Explanation:

Given;

length of steel wire, L = 100 ft

cross-sectional area, A = 0.0144 in²

applied force, F = 270 lb

extension of the wire, e = 0.75 in

<u>Part (A)</u> The stress on the steel wire;

δ = F/A

   = 270 / 0.0144

δ  = 18750 lb/in² = 19,000 Psi

<u>Part (B)</u> The strain on the steel wire;

σ = e/ L

L = 100 ft = 1200 in

σ = 0.75 / 1200

σ = 0.00063

<u>Part (C)</u> The modulus of elasticity of the steel

E = δ/σ

   = 19,000 / 0.00063

E = 30,000,000 Psi

4 0
4 years ago
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