An interest rate of norminal 12% per year , compounded weekly is

If a hoist lifts a 4500lb load 30ft in 15s, the power delivered to the load is a) 18.00hp b) 9000hp c) 16.36hp d) None of the ab

12345 [234]

Answer:

Explanation:

load = 4500lb lift height= 30 ft

time =15 s

velocity= $\frac{30}{15}$ ft/s

velocity=2 ft/s

power = force $\times$ velocity

power= ${4500}\times2$

power= 9000 lb ft/s

1 hp= 550 lb ft/s

power= $\frac{9000}{550} =16.36$ hp

5 0

3 years ago

A contractor is planning on including several skylights in each unit of a residential development. What type of worker would she

Sladkaya [172]

Answer:

Glazier

Explanation:

Glaziers are workers who specializes in cutting and installation of glass works.

They work with glass in various surfaces and settings, such as cutting and installing windows and doors, skylights, storefronts, display cases, mirrors, facades, interior walls, etc.

Thus, the type of worker the contractor will hire for this project is a Glazier

8 0

3 years ago

An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at

Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

$L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft$

$G_2$ is given as

$G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%$

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

$A=\frac{L}{K}\\A=\frac{460}{115}\\A=4$

A is given as

$-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%$

With initial grade, the elevation of PVC is

$E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\$

The station is given as

$St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\$

Low point is given as

$x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft$

The station of low point is given as

$St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\$

The elevation is given as

$E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft$

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

3 0

3 years ago

1.8 A water flow of 4.5 slug/s at 60 F enters the condenser of steam turbine and leaves at 140 F. Determine the heat transfer ra

Ann [662]

Answer:

$Hr=4.2*10^7\ btu/hr$

Explanation:

From the question we are told that:

Water flow Rate $R=4.5slug/s=144.78ib/sec$

Initial Temperature $T_1=60 \textdegree F$

Final Temperature $T_2=140 \textdegree F$

Let

Specific heat of water $\gamma= 1$

And

$\triangle T= 140-60$

$\triangle T= 80\ Deg.F$

Generally the equation for Heat transfer rate of water $H_r$ is mathematically given by

Heat transfer rate to water= mass flow rate* specific heat* change in temperature

$H_r=R* \gamma*\triangle T$

$H_r=144.78*80*1$

$H_r=11582.4\ btu/sec$

Therefore

$H_r=11582.4\ btu/sec*3600$

$Hr=4.2*10^7\ btu/hr$

3 0

3 years ago

Base course aggregate has a target dry density of 119.7 Ib/cu ft in place. It will be laid down and compacted in a rectangular s

djyliett [7]

Answer:

total weight of aggregate = 5627528 lbs = 2814 tons

Explanation:

we get here volume of space to be filled with aggregate that is

volume = 2000 × 48 × 0.5

volume = 48000 ft³

now filling space with aggregate of the density that is

density = 0.95 × 119.7

density = 113.72 lb/ft³

and dry weight of this aggregate is

dry weight = 48000 × 113.72

dry weight = 5458320 lbs

we consider here percent moisture is by weigh

so weight of moisture in aggregate will be

weight of moisture = 0.031 × 5458320

weight of moisture = 169208 lbs

so here total weight of aggregate is

total weight of aggregate = 5458320 + 169208

total weight of aggregate = 5627528 lbs = 2814 tons

3 0

3 years ago