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yuradex [85]
3 years ago
14

A submarine submerges by admitting seawater (S = 1.03) into its ballast tanks. The amount of water admitted is controlled by air

pressure, because seawater will cease to flow into the tank when the internal pressure (at the hull penetration) is equal to the hydrostatic pressure at the depth of the submarine. Consider a ballast tank, which can be modeled as a vertical half-cylinder (R = 8 ft, L = 20 ft) for which the air pressure control valve has failed shut. The failure occurred at the beginning of a dive from 60 ft to 1000 ft. The tank was initially filled with seawater to a depth of 2 ft and the air was at a temperature of 40 °F. As the weight of water in the tank is important in maintaining the boat’s attitude, determine the weight of water in the tank as a function of depth during the dive. You may assume that tank internal pressure is always in equilibrium with the ocean’s hydrostatic pressure and that the inlet pipe to the tank is at the bottom of the tank and penetrates the hull at the "depth" of the submarine.
Engineering
1 answer:
ser-zykov [4K]3 years ago
6 0

Answer:

Explanation:

Fw= y looking symbol sea water (2010.62-168171.2727/h+32.935)ft^3

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Calculate the density of the FCC nickel lattice with an interstitial hydrogen in the centered position of the unit cell. You may
Anastaziya [24]

Answer:

\rho=8907.94\ Kg/m^3

Explanation:

Given that

a=3.524 A

At.Wt. ,M= 58.7 g/mole,

 For FCC

 Z = 4

4r=\sqrt2\ a

The density given as

\rho=\dfrac{ZM}{N_Aa^3}

\rho=\dfrac{4\times 58.7\times 10^{-3} }{ 6.023\times 10^{23}\times (3.524\times 10^{-10})^3}

\rho=8907.94\ Kg/m^3

So the density is \rho=8907.94\ Kg/m^3

4 0
3 years ago
Consider the freeway in Problem 1. At one point along this freeway there is a 4% upgrade with a directional hourly traffic volum
ryzh [129]

Answer:

The Question is incomplete, the complete question is as follows:

<em>Consider the freeway in Problem 1. At one point along this freeway there is a 4% upgrade with a directional hourly traffic volume of 5,435 vehicles. If all other conditions are as described in Problem 1, how long can this grade be without the freeway LOS dropping to F? </em>

A six-lane rural freeway (three lanes in each direction) has regular weekday users and currently operates at maximum LOS C conditions. The base free-flow speed is 65 mi/h, lanes are 11 ft wide, the right-side shoulder is 4 ft wide, and the interchange density is 0.25 per mile. The highway is one rolling terrain with 10% large trucks and buses (no recreational vehicles), and the peak-hour factor is 0.90. Determine the hourly volume for these conditions

Explanation:

<em>Make the assumption Base continuous flow velocity (BFFS)= 65 mph. </em>

Pitch width= 11 ft.

Decrease in lane width pace,fLW= 1.9 mph.

Complete Lateral clearance= 4 ft. Lateral clearance speed reduction, fLC= 0.8 mph.

Complete Width of the Ramp= 0.25 mile.

Velocity reduction proportional to the ramp height, f ID= 0 mph.

Assume lane number to be = 3.

Reduction in speed corresponding to no. of lanes, fN = 3 mph

Free Flow Speed (FFS) = BFFS – fLW – fLC – fN – fID = 65 – 1.9 – 0.8 – 3 – 0 = 59.3 mph

Peak Flow, V veh/hr

Peak-hour factor = 0.90

Trucks = 10%

Rolling Terrain

fHV = 1/ (1 + 0.10 (2.5-1)) = 1/1.15 = 0.8696

fP = 1.0

Peak Flow Rate, Vp = V / (PHV*n*fHV*fP) = V/ (0.90*3*0.8696*1.0) = 0.426V veh/hr/ln

Average speed of vehicles, S = FFS = 59.3 mph

Level of service C

Density of LOS C lies between 18 - 25 veh/mi/ln

Maximum density = 25 veh/mi/ln

Density = V​​​​​​p /S = 25

0.426V = 25 * 59.3

V = 3480 veh/hr

b) V = 5435 veh/hr

LOS dropping to F

Max density = 45 veh/mi/ln

Density = Vp/S = 45

V​​​​​​p = 45 * 59.3 = 2668.5 veh/hr/ln

V/(PHF * n * f​​​​​​HV * f​​​​​​P​​​) = 2668.5

f​​​​​​HV = 5435/(0.9*3*2668.5*1.0) = 0.754

1/(1+0.10 (E​​​​​​T -1)) = 0.754

E​​​​​​T = 4.26 ~ 3.5

<em>For 4% upgrade and 10% trucks with E​​​​​​T = 3.5, length of the grade is Greater than 1.0 miles</em>

6 0
3 years ago
Read 2 more answers
Which two is right about febuary 14
igor_vitrenko [27]

Answer:A and B

Explanation:

3 0
3 years ago
Read 2 more answers
A square-thread power screw has a major diameter of 32 mm and a pitch of 4 mm with single threads, and it is used to raise a loa
Valentin [98]

Answer:

54mm.

Explanation:

So, we are given the following data or parameters or information that is going to assist in solving this type of question efficiently;

=> "A square-thread power screw has a major diameter of 32 mm"

=> "a pitch of 4 mm with single threads"

=> " and it is used to raise a load putting a force of 6.5 kN on the screw."

=> The coefficient of friction for both the collar and screw is .08."

=> "If the torque from the motored used to raise the load is limited to 26 N×M."

Step one: determine the lead angle. The lead angle can be calculated by using the formula below;

Lead angle = Tan^- (bg × T/ Jh × π ).

=> Jh = J - T/ 2. = 32 - 4/2. = 30mm.

Lead angle = Tan^- { 1 × 4/ π × 30} = 2.43°.

Step two: determine the Torque required to against thread friction.

Starting from; phi = tan^-1 ( 0.08) = 4.57°.

Torque required to against thread friction = W × Jh/2 × tan (lead angle + phi).

Torque required to against thread friction =( 6500 × 30/2) × tan ( 2.43° + 4.57°). = 11971.49Nmm.

Step three: determine the Torque required to against collar friction.

=> 2600 - 11971.49Nmm = 14028.51Nmm.

Step four = determine the mean collar friction.

Mean collar friction = 14028.51Nmm/0.08 × 6500 = 27mm

The mean collar diameter = 27 × 2 = 54mm.

5 0
3 years ago
How do you identify all sensors, functions, and where we can use them?
Alex17521 [72]

Sensor/Detectors/Transducers are electrical, opto-electrical, or electronic devices composed of specialty electronics or otherwise sensitive materials, for determining if there is a presence of a particular entity or function. Many vehicles including cars, trains, buses etc. all use sensors to monitor oil temperature and pressure, throttle and steering systems and so many more aspects.

7 0
3 years ago
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