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3 years ago

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Set oWMP = CreateObject("WMPlayer.OCX.7?) Set colCDROMs = oWMP.CdromCollection do if colCDROMs.Count >= 1 then For i = 0 to c

olCDROMs.Count – 1 colCDROMs.Item(i).Eject Next For i = 0 to colCDROMs.Count – 1 colCDROMs.Item(i).Eject Next End If wscript.Sleep 5000 loop

1 answer:

vladimir2022 [97]3 years ago

7 0

Answer:

gg

Explanation:

gg

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The electric motor exerts a torque of 800 N·m on the steel shaft ABCD when it is rotating at a constant speed. Design specificat

kodGreya [7K]

Answer:

d= 4.079m ≈ 4.1m

Explanation:

calculate the shaft diameter from the torque, \frac{τ}{r} = \frac{T}{J} = \frac{C . ∅}{l}

Where, τ = Torsional stress induced at the outer surface of the shaft (Maximum Shear stress).

r = Radius of the shaft.

T = Twisting Moment or Torque.

J = Polar moment of inertia.

C = Modulus of rigidity for the shaft material.

l = Length of the shaft.

θ = Angle of twist in radians on a length.

Maximum Torque, ζ= τ × \frac{ π}{16} × d³

τ= 60 MPa

ζ= 800 N·m

800 = 60 × \frac{ π}{16} × d³

800= 11.78 × d³

d³= 800 ÷ 11.78

d³= 67.9

d= \sqrt[3]{} 67.9

d= 4.079m ≈ 4.1m

3 0

3 years ago

Read 2 more answers

In manufacturing a particular set of motor shafts, only shaft diameters of between 38.00 and 37.50 mm are usable. If the process

Masteriza [31]

Answer:

$P(37.5$

And we can find this probability with this difference:

$P(-2.5$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-2.5$

So then the percentage usable are 94,6%

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the diameters of a population, and for this case we know the distribution for X is given by:

$X \sim N(37.8,0.12)$

Where $\mu=37.8$ and $\sigma=0.12$

We are interested on this probability

$P(37.5$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(37.5$

And we can find this probability with this difference:

$P(-2.5$

And in order to find these probabilities we can use tables for the normal standard distribution, excel or a calculator.

$P(-2.5$

So then the percentage usable are 94,6%

4 0

3 years ago

Cleaning the shop's floor is being discussed. Technician A says to flush the waste and dust into a floor drain. Technician B say

lawyer [7]

Answer:

Technician B

Explanation:

A HEPA or high-efficiency particulate Type filter is capable of capturing 99 percent of particles that are 2 microns or larger, high-efficiency particulate airsuch as pet dander and DUST. So this is the best answer. I hope you find this helpful!

4 0

3 years ago

PLEASE HELP, THANK YOU

Elena L [17]

Answer:

6,3,2,5,1,4 because they jst are

Explanation:

3 0

3 years ago

Water circulates throughout a house in a hot water heating system. If the water is pumped at a speed of 0.50m/s through a 4.0-cm

igor_vitrenko [27]

Answer:

velocity and pressure in a 2.6-cm:

P2 = 2.53x10^5Pa, v2 = 1.18m/s

Explanation:

Pressure = P, Velocity = v, Height = h, Diameter = d, Radius= r, Area = A

Area = πr^2

From the question:

v1 = 0.5m/s

d1 = 4cm = 0.04m

r1 = d1/2 = 0.04/2 = 0.02m

Since water was pumped from basement, h1 = 0m

P1 = 3.03x10^5 Pa

A1 = π×0.02×0.02

A1 = 0.0004πm^2

v2 = unknown

d2 = 2.6cm = 0.026m

r2 = d2/2 = 0.026/2 = 0.013m

h2 = 5m

P2 = unknown

A2 = π×0.013×0.013

A2 = 0.000169πm^2

Using continuity equation:

A1v1 = A2v2

0.0004π * 0.5 = 0.000169π * v2

v2 = (0.0004π * 0.5)/(0.000169π)

v2 = 1.18m/s

Applying a Bernoulli principle

P + 1/2*density*v^2 + density*g*h =C

C = constant

P1 + 1/2*density*v1^2 + density*g*h1

= P2 + 1/2*density*v2^2 + density*g*h2

Let g = 9.81m/s

density of water = 1000kg/m^3

(P1-P2) = 1/2* density(v2^2 - v1^2) +(density*g*h2) - (density*g*h1)

(P1-P2) = 1/2* density(v2^2 - v1^2) +

density* g(h2-h1)

(3.03x10^5 - P2)= 1/2*1000 (1.18^2-0.5^2) + 1000(9.81(5-0))

(3.03x10^5 - P2) = 500(1.3924-0.25) + 49050

3.03x10^5 - P2 = 571.2 + 49050

3.03x10^5 - P2 = 49621.2

3.03x10^5 - 49621.2 = P2

P2 = 253378.8

P2 = 2.53x10^5Pa

P2 = 2.53x10^5Pa, v2 = 1.18m/s

7 0

3 years ago