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3 years ago

8

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

ove

1 answer:

12345 [234]3 years ago

5 0

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

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A hemispherical shell with an external diameter of 500 mm and a thickness of 20 mm is going to be made by casting, located entir

Olenka [21]

Solution :

Given :

External diameter of the hemispherical shell, D = 500 mm

Thickness, t = 20 mm

Internal diameter, d = D - 2t

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Density of molten metal, ρ = $7.2 \ g/cm^3$

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The height of pouring cavity above parting surface is h = 300 mm

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Area, A $=2 \pi r^2$

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$=7200 \times 9.81 \times 0.3324 \times 0.3$

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2 years ago

Technician A ay that acid rain doe the mot harm when it firt fall on a finih. Technician B ay that hard water potting i uually j

Tamiku [17]

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1 year ago

A rigid tank whose volume is 2 m3, initially containing air at 1 bar, 295 K, is connected by a valve to a large vessel holding a

bazaltina [42]

Answer:

$Q_{cv}=-339.347kJ$

Explanation:

First we calculate the mass of the aire inside the rigid tank in the initial and end moments.

$P_iV_i=m_iRT_i$ (i could be 1 for initial and 2 for the end)

State1

$1bar*|\frac{100kPa}{1}|*2=m_1*0.287*295$

$m_1=232kg$

State2

$8bar*|\frac{100kPa}{1bar}|*2=m_2*0.287*350$

$m_2=11.946$

So, the total mass of the aire entered is

$m_v=m_2-m_1\\m_v=11.946-2.362\\m_v=9.584kg$

At this point we need to obtain the properties through the tables, so

For Specific Internal energy,

$u_1=210.49kJ/kg$

For Specific enthalpy

$h_1=295.17kJ/kg$

For the second state the Specific internal Energy (6bar, 350K)

$u_2=250.02kJ/kg$

At the end we make a Energy balance, so

$U_{cv}(t)-U_{cv}(t)=Q_{cv}-W{cv}+\sum_i m_ih_i - \sum_e m_eh_e$

No work done there is here, so clearing the equation for Q

$Q_{cv} = m_2u_2-m_1u_1-h_1(m_v)$

$Q_{cv} = (11.946*250.02)-(2.362*210.49)-(295.17*9.584)$

$Q_{cv}=-339.347kJ$

The sign indicates that the tank transferred heat<em> to</em> the surroundings.

8 0

2 years ago

How should you move your board through the planer? (Pick two choices.)

iragen [17]

Answer:

I would say do it at an even pace

Explanation:

Doing it a slow pace takes time quickly will probably not to good gor you and doing it at an irregular pace is just way to fast

4 0

3 years ago

I will mark as brainliest !

Sliva [168]

Answer:

7.8 Mph

Explanation:

Rate of cycling = 1.1 rev/s

Rear wheel diameter = 26 inches

Diameter of sprocket on pedal = 6 inches

Diameter of sprocket on rear wheel = 4 inches

Circumference of rear wheel = \pi d=26\piπd=26π

Speed would be

\begin{gathered}\text{Rate of cycling}\times \frac{\text{Diameter of sprocket on pedal}}{\text{Diameter of sprocket on rear wheel}}\times{\text{Circumference of rear wheel}}\\ =1.1\times \frac{6}{4}\times 26\pi\\ =134.77432\ inches/s\end{gathered}Rate of cycling×Diameter of sprocket on rear wheelDiameter of sprocket on pedal×Circumference of rear wheel=1.1×46×26π=134.77432 inches/s

Converting to mph

1\ inch/s=\frac{1}{63360}\times 3600\ mph1 inch/s=633601×3600 mph

134.77432\ inches/s=134.77432\times \frac{1}{63360}\times 3600\ mph=7.65763\ mph134.77432 inches/s=134.77432×633601×3600 mph=7.65763 mph

The Speed of the bicycle is 7.8 mph

3 0

3 years ago