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3 years ago

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Two identical 3 in. major-diameter power screws (single-threaded) with modified square threads are used to raise and lower a 50-

ton sluice gate of a dam. Quality of construction and maintenance (including lubrication) are good, resulting in an estimated friction coefficient of only 0.1 for the screw. Collar friction can be neglected, as ball thrust bearings are used. Assuming that, because of gate friction, each screw must provide a lifting force of 26 tons, what power is required to drive each screw when the gate is being raised at the rate of 3 ft/min

1 answer:

sp2606 [1]3 years ago

6 0

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

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A hawser is wrapped two full turns around a bollard. By exerting an 80-lb force on the free end of the hawser, a dockworker can

Brut [27]

Answer:

μ=0.329, 2.671 turns.

Explanation:

(a) ln(T2/T1)=μβ β=angle of contact in radians

take T2 as greater tension value and T1 smaller, otherwise the friction would be opposite.

T2=5000 lb and T1=80 lb

we have two full turns which makes total angle of contact=4π radians

μ=ln(T2/T1)/β=(ln(5000/80))/4π

μ=0.329

(b) using the same relation as above we will now compute the angle of contact.

take greater tension as T2 and smaller as T1.

T2=20000 lb T1=80 lb μ=0.329

β=ln(20000/80)/0.329=16.7825 radians

divide the angle of contact by 2π to obtain number of turns.

16.7825/2π =2.671 turns

4 0

3 years ago

A pipe produces successive harmonics at 300 Hz and 350 Hz. Calculate the length of the pipe and state whether it is closed at on

MAXImum [283]

Answer:

The pipe is open ended and the length of pipe is 3.4 m.

Explanation:

For identification of the type of pipe checking the successive frequencies in both the open pipe and closed pipe as below

Equation for nth frequency for open end pipe is given as

$f_n=\frac{nv}{2L}$

For (n+1)th value the frequency is

$f_{n+1}=\frac{(n+1)v}{2L}$

Taking a ratio of both equation and solving for n such that the value of n is a whole number

$\frac{f_{n+1}}{f_n}=\frac{\frac{(n+1)v}{2L}}{\frac{nv}{2L}}\\\frac{350}{300}=\frac{(n+1)}{n}\\350n =300n+300\\50n =300\\n =6\\$

So n is a whole number this means that the pipe is open ended.

For confirmation the nth frequency for a closed ended pipe is given as

$f_n=\frac{(2n+1)v}{4L}$

For (n+1)th value the frequency is

$f_{n+1}=\frac{(2n+3)v}{4L}$

Taking a ratio of both equation and solving for n such that the value of n is a whole number

$\frac{f_{n+1}}{f_n}=\frac{\frac{(2n+3)v}{2L}}{\frac{(2n+1)v}{2L}}\\\frac{350}{300}=\frac{(2n+3)}{(2n+1)}\\700n+350 =600n+900\\100n =550\\n =5.5\\$

As n is not a whole number so this is further confirmed that the pipe is open ended.

Now from the equation of, with n=6, v=340 m/s and f=300 Hz

$f_n=\frac{nv}{2L}\\300=\frac{6 \times 340}{2L}\\L=\frac{2040}{600}\\L=3.4 m$

The value of length is 3.4m.

5 0

3 years ago

Design a ductile iron pumping main carrying a discharge of 0.35 m3/s over a distance of 4 km. The elevation of the pumping stati

snow_tiger [21]

Answer:

$D=0.41m$

Explanation:

From the question we are told that:

Discharge rate $V_r=0.35 m3/s$

Distance $d=4km$

Elevation of the pumping station $h_p= 140 m$

Elevation of the Exit point $h_e= 150 m$

Generally the Steady Flow Energy Equation SFEE is mathematically given by

$h_p=h_e+h$

With

$P_1-P_2$

And

$V_1=V-2$

Therefore

$h=140-150$

$h=10$

Generally h is give as

$h=\frac{0.5LV^2}{2gD}$

$h=\frac{8Q^2fL}{\pi^2 gD^5}$

Therefore

$10=\frac{8Q^2fL}{\pi^2 gD^5}$

$D=^5\frac{8*(0.35)^2*0.003*4000}{3.142^2*9.81*10}$

$D=0.41m$

8 0

3 years ago

NEED HELPASAP what is the moisture content of a board if a test sample that originally weighed 11. 5 oz was found to weigh 10 oz

Viefleur [7K]

Answer:

11.5-10/10= 0.15x100 =15%

Explanation:

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2 years ago

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inna [77]

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3 years ago