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lora16 [44]
2 years ago
15

suppose a wheel with a 15 inch diameter is used to turn a water valve stem with a radius of .95 inches. What is the Mechanical a

dvantage?
Engineering
1 answer:
pentagon [3]2 years ago
5 0

Answer: The answer is 7.89

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Spring-loaded rack guide yokes are made of ______ and have a spring that pushes on the back side of the rack to help reduce the
Kobotan [32]

Answer:

metal

Explanation:

A yoke assembly are use in an assembly of a rack and pinion steering gear for a vehicle. The spring loaded yokes guided with a rack are made up of metals. It consists of a spring made of steel which pushes the back side of the rack to reduce the play that occurs between the pinion and the rack and still allow the relative motion.

6 0
3 years ago
A lake with an area of 525 [acre] was monitored during a one-month period. The average inflow was 30 [cfs] for the month and the
Alex73 [517]

Answer:

\Delta S = 1581663.5ft^3

Explanation:

We need to calculate the change in storage through the changes given,

That is,

\Delta S = P+I-O-O_{seepage}-E

Where the loss are representing by,

P= Precipitation\\I= Inflow\\O= Outflow\\O_{Seepage}= Outflow by seepage\\E=Evaporation

So calculating the values we have

\Delta S = P+(I-0)-O_{seepage}-E

\Delta S = 4.25+ (30ft^3/s-27ft^3/s)-1.5in-6in

The values inside the are parenthesis need to be konverted as I note here.

(30days(24hr/1day)(3600s/1hr)(1acre.ft/43560ft^3)(1/525acres)(12in/1ft)

That is,

\Delta S = 0.83in\Delta S= (0.83in*1ft/12in)(525acres)\\\Delta S=36.31 acres.ft\\\Delta S=36.31acres.ft*(43560ft^3/acrees.ft)\\\Delta S = 1581663.5ft^3

3 0
3 years ago
Make Routh tables and tell how many roots of the following two polynomials are in the right half plane and in the left half plan
12345 [234]

Answer:  

P(s) = s5 + 3s4 + 5s3 + 4s2 + s

For above Polynomial, there are two RHP Poles & Three LHP Poles.

& for the Second Polynomial,

P(s)=3s7 + 9s6 + 6s5 + 4s4 + 7s3 + 8s2 + 2s + 6

For this Polynomial, there are Four RHP Poles & Three LHP Poles.

Explanation:

As the equation is given by for the first polynomial,

P(s)=s5 + 3s4 + 5s3 + 4s2 + s

The Routh table is attached in the attachment, & from that table we can see that there are two sign changes for thuis first ploynomial. Moreover, there are two RHP poles & Three LHP Poles.

Similarly the equation for the second polynomial is given by as,

P(s)=3s7 + 9s6 + 6s5 + 4s4 + 7s3 + 8s2 + 2s + 6

The Routh Table is attached in the same attachment below in the 2nd slide & from this table we can see that there are four sign changes for this polynomial while it includes Four RHP Poles & three LHP Poles.

Download pptx
3 0
3 years ago
The following information pertains to the January operating budget for Casey Corporation. ∙ Budgeted sales for January $200,000
stich3 [128]

Answer:

For January, budgeted cost of goods sold is $150,000

Explanation:

Sales is the consists of Gross income and Cost of goods sold. We can derive the gross income value after deducting cost of goods sold from sales value.

Budgeted Sales for next month = $200,000

Gross Margin = 25%

Gross Margin = Gross Income / Sales

25% = Gross Income / $200,000

Gross Income is the net value of sales and cost of goods sold.

Gross Income = $200,000 x 25%

Gross Income = $50,000

As we know,

Gross Income = Sales - Cost of Goods sold

$50,000 = $200,000 - Cost of Goods sold

Cost of Goods sold = $200,000 - $50,000

Cost of Goods sold = $150,000

5 0
3 years ago
Soil is to be excavated from a borrow pit to construct a 100-m-long trapezoidal levee that is 3 meters high, 5 meters wide at th
shepuryov [24]

Answer:

The answer is 2217.32m³

Explanation:

Given that:

V₂  = (15 + 5/2) * 3 * 100 = 3000 m³

r₁ = 21.7kN/m3

w (water) = 11.4%.

e₂ (void ratio) = 0.47

V₁ = ?

Now,

rd₁ =  r₁/1 + w₁ = 21.7/1 +(11.4/100)

= 19.47 kN/m3

Thus,

rd₂ = r₁/1 + e₂ = 21.7/1+ 0.47 = 14.76 kN/m3

Then,

V₁/V₂ = rd₁/ rd₂

V₁ = 3000 * 14.76/19.97

V₁ = 2217.32m³

Therefore, volume that should be excavated from the borrow pit is 2217.32m³

4 0
3 years ago
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