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Daniel [21]
3 years ago
6

A masonry facade consisting of 3,800 square feet is to be constructed for a building. The total cost per worker hour is estimate

d to be $31.50, and the total estimated cost of the labor for this task is $10,500. Assuming 8-hour work days and a crew of six workers, how many days should be allowed to complete this task? What is the production rate that the crew must attain to keep the project on schedule and within the budget?
Engineering
1 answer:
lana66690 [7]3 years ago
6 0

Answer:

Days: 6.9444 days

Production rate: 547.2035 ft²/s

Explanation:

the solution is attached in the Word file

Download docx
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"Write a statement that outputs variable numItems. End with a newline. Program will be tested with different input values."
kirill [66]

Answer:

The solution code is written in Java.

System.out.println(numItems);

Explanation:

Java <em>println() </em>method can be used to display any string on the console terminal. We can use <em>println()</em> method to output the value held by variable <em>numItems.</em> The <em>numItems </em>is passed as the input parameter to <em>println()</em> and this will output the value of <em>numItems</em> to console terminal and at the same time the output with be ended with a newline automatically.  

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Describe the cycle that purifies drinking water. 40 points
Sunny_sXe [5.5K]
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3 years ago
Read 2 more answers
Can you help me with this
Fiesta28 [93]

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6 0
2 years ago
A 3-ft-diameter vertical cylindrical tank open to the atmosphere contains 1-ft-high water. The tank is now rotated about the cen
arlik [135]

Answer:

The angular velocity is 7.56 rad/s

the maximum water height is 2 ft

Explanation:

The z-position as a function of r is equal to

z_{s(r)} =h_{0} -\frac{w^{2}(R^{2}-2r^{2}   }{4g} (eq. 1)

where

h0 = initial height = 1 ft

w = angular velocity

R = radius of the cylinder = 1.5 ft

zs(r) = 0 when the free surface is lowest at the centre

Replacing and clearing w

w=\sqrt{\frac{4gh_{0} }{R^{2} } } =\sqrt{\frac{4*32.17*1}{1.5^{2} } } =7.56rad/s

if you consider the equation 1 for the free surface at the edge is equal to

z_{s(R)} =h_{0} +\frac{w^{2}R^{2}   }{4g} =1+\frac{(7.56^{2})*(1.5^{2} ) }{4*32.17} =1.99ft=2ft

7 0
3 years ago
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