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miv72 [106K]
3 years ago
15

Specify whether the statements are true or false.

Engineering
1 answer:
NISA [10]3 years ago
6 0

Answer:

1. False

2. True

3. True

4. True

5. False

Explanation:

Moment of a force is not a free vector. There are certain quantities along the line with which force is applied.

Force can be moved in any direction along the line of the action without changing the external reaction.

The magnitude of equivalent resultant force is distributed along the centroid point.

The resultant force of a couple force system is zero as it form opposite forces which balances off each other.

You might be interested in
Explain why change is inevitable in complex systems and give examples (apart from prototyping and incremental delivery) of softw
Over [174]

Explanation:

The change in complex systems can be explained according to the relationship of the environment where the system is implemented.

The system environment is dynamic, which consequently leads to adaptation to the system, which generates new requirements inherent to changes in business objectives and policies. Therefore, changing systems is necessary for tuning and usefulness so that the system correctly supports business requirements.

An example is the registration of the justification of the requirements, which is a process activity that supports changes in the system so that the reason for including a requirement is understood, which helps in future changes.

8 0
3 years ago
Liquid water is fed to a boiler at 24°C and 10 bar is converted at a constant pressure to saturated steam.
zepelin [54]

We can find the change in the enthalpy through the tables A5 for Saturated water, pressure table.

For 1bar=1000kPa:

T_{sat}=179.88\°c

H_{fg} = 2014.6kJ/kg

c_p=4.18 kJkg^{-1}{K^{-1}

\nu_g = 0.19436m^3/kg

Replacing,

\Delta h = h_{fg}+c_p(T_{sat}-T_{inlet})

\Delta h = 2014.6+4.18(179.88-24)

\Delta h=2666.17kJ/kg

With the specific volume we know can calculate the mass flow, that is

\dot{m}=\frac{\frac{15000}{3600}}{0.19436}

\dot{m} = 21.4378kg/s

Then the heat required in input is,

Q=\dot{m}\Delta h

Q=21.4378*2666.17

Q=57157.036kW

With the same value required of 15000m^3/h, we can calculate the velocity of the water, that is given by,

V= \frac{\dotV}{A}

V = \frac{\frac{15000}{3600}}{\pi /4 *(0.15)^2}

V=235.79m/s

Finally we can apply the steady flow energy equation, that is

\dot{m}(h_1+\frac{V^2}{2000})+Q = \dot{m}h_2

Re-arrange for Q,

Q=\dot{m}(h_2-h_1-\frac{V^2}{2000})

Q=\dot{m}(\Delta h-\frac{V^2}{2000})

Q= (21.4378)(2666.17-\frac{235.79^2}{2000})

Q= 56560.88kW

We can note that consider the Kinetic Energy will decrease the heat input.

4 0
3 years ago
If you should lose your balance, you should grab onto the turning center to steady yourself.
lisov135 [29]

No/false is the answer

6 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
You have been assigned to design an open cylindrical storage tank 4 meters tall with a diameter of 8 meters to be made out of A-
Katen [24]

Answer:

The required wall thickness is 1.506 \times 10^{-3} m

Explanation:

Given:

Fluid density \rho = 1200 \frac{kg}{m^{3} }

Diameter of tank d = 8 m

Length of tank l = 4 m

F.S = 4

For A-36 steel yield stress \sigma = 250 MPa,

Allowable stress \sigma _{allow} = \frac{\sigma}{F.S}

 \sigma _{allow} = \frac{250}{4} = 62.5 MPa

Pressure force is given by,

 P = \rho gh

 P = 1200 \times 9.8 \times 4

P = 47088 Pa

Now for a vertical pipe,

\sigma _{allow} = \frac{Pd}{4t}

Where t = required thickness

 t = \frac{Pd}{4 \sigma _{allow} }

 t = \frac{47088 \times 8 }{4 \times 62.5 \times 10^{6} }

t = 1.506 \times 10^{-3} m

Therefore, the required wall thickness is 1.506 \times 10^{-3} m

8 0
3 years ago
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