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2 years ago

Which statement concerning symbols used on plans is true?

Engineering

1 answer:

Zielflug [23.3K]2 years ago

8 0

Answer:

The symbols are to be noted on the title sheet or other introductory sheet of the plans.

Explanation:

in able to be understood for example a map key is always on a map so the reader can use it efficently without confusion

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What is the air change rate (ACH) for a 100 ft^2 (9.3 m^2) space with a 10 ft (3.0 m) ceiling and an airflow rate of 200 cfm (95

kakasveta [241]

Answer:

The ACH is 12/h

Solution:

As per the question:

Area of the space, $A_{s} = 100 ft^{2}$

Height of the given space, h = 10 ft

Air flow rate, $Q_{a} = 200 cfm$

Now, to find the Air Change Rate (ACH):

We calculate the Volume of the given space:

$V_{s} = A_{s}\times h = 100\times 10 = 1000 ft^{3}$

Now, the ACH per min:

= $\frac{V_{s}}{Q_{a}} = \frac{1000}{200} = 5/min$

Now, ACH per hour:

= $\frac{60}{5} = 12/h$

3 0

3 years ago

A Michelson interferometer operating at a 500 nm wavelength has a 3.73-cm-long glass cell in one arm. To begin, the air is pumpe

LekaFEV [45]

Answer:

The number of bright-dark fringe is 42

Solution:

As per the question:

Wavelength of light, $\lambda = 500\ nm = 500\times 10^{- 9}\ m$

Length of the glass cell, x = 3.73 cm = 0.0373 m

Refractive index, $\mu = 1.00028$

Now,

To calculate the bright-dark fringe shifts, we use the formula given below:

$d_{m} = \frac{2x}{\lambda }\times (\mu - 1)$

Now, substituting the appropriate values in the above formula:

$d_{m} = \frac{2\times 0.0373}{500\times 10^{- 9}}\times (1.00028 - 1)$

$d_{m} = 41.77$ ≈ 42

6 0

3 years ago

Which symbol should be used for the given scenario?

Firdavs [7]

Answer:

speed square

Explanation:

4 0

2 years ago

The two boxcars A and B have a weight of 20000lb and 30000lb respectively. If they coast freely down the incline when the brakes

kolbaska11 [484]

Answer:

T=5.98 kips

Explanation:

First, introduce forces, acting on both cars:

on car A there are 4 forces acting: gravity force mA*g, normal reaction force, friction force and force T- it represents the interaction between cars A and B. On car B, there are three forces acting: gravity force, normal reaction force and force T. Note, that force T is acting on both cars, but it has opposite direction. Force T, acting on car A has direction, opposite to the friction force, whether the T, acting on B, is directed backwards- in the same direction with the friction force. Note, that both cars have the same acceleration, which is directed backwards.

Once the forces were established, we can write components of the Second Newtons Law on vertical and horizontal axes, considering that horizontal axis is directed backwards- in the same direction with the acceleration:

For car A on the vertical axis the equation is: -mAg+NA=0

For car A on the horizontal axis, the equation is: Ffr-T=mAa

For car B, on the vertical axis the equation is: -mBg+NB=0

For car B, on the horizontal axis, the equation is: T=mBa

We need to solve these equations to find force T, knowing that Ffr=μmAg, where

After the transformations, the equations for acceleration and force in the coupling will be:

a=(μmAg)/(mA+mB)=6.43 ft/s2- note, that the given answer is not correct for the given numerical values;

and force T: T=μmAmBg/(mA+mB)=6.0 kips- note, that the force answer is in line with the given numerical value

5 0

3 years ago

. Consider the single-engine light plane described in Prob. 2. If the specific fuel consumption is 0.42 lb of fuel per horsepowe

Trava [24]

Answer:

Hence the Range and Endurance of single engine plane is given by

650.644 miles and 5.3528 hrs at standard sea level.

Explanation:

Given :

A single engine light plane with ,

Specific fuel consumption 0.42lb/hr/hp.

Fuel capacity =44 gal.

Gross weight =3400 lb.

To find :

Range and Endurance of the plane.

Solution:

Consider all standard measures of standard single engine propeller plane

Wing span =35.8 fts.

Wing swing area=174 sq ft

parasite drag coefficient =Cd.o.=0.025

Oswald's eff. factor= 0.8

ρ=0.002377= corresponds to standard sea level constant.

Now

Formula for Range is given by, Breguent formula.

R=(η/c) *(Cl/Cd)*ln(W1/W0)

here η is Oswald's constant,

Now calculating lift(Cl) and drag coefficient (Cd)

Cl=W/(1/2*ρ*v^2*S)

W=Gross weight

ρ=0.002377

Assume v=200 ft/sec normally,

S=174 Sq .ft.

CI=3400/(1/2*0.002377*200*200*174)

=6800/16543.9

=0.4110

Now calculating drag constant,

AR=(wing span)^2/wing swing area

=(35.8)^2/174

=7.37

Now

Drag Coefficient

Cd=Cd.o.+ (Cl^2)/(pie*e*AR)

=0.025+(0.4110)^2/(3.142*0.8*7.36)

=0.0342

Given that 44 gal fuel capacity and in Aviation weight of fuel is 5.64 lb/gal

hence weight of fuel=W1=3400- (44*5.64)

=3151.84

Now

for specific fuel consumption=0.42 lb/hp/hr

=0.42 lb*(1/550 ft)*(1/3600)sec

=2.12 *10^-7 lb/ft/sec

Now further calculating range

R=(η/c) *(Cl/Cd)*ln(W1/W0)

={0.8/(2.12*10^-7)}*(0.4110/0.0342)*ln(3151.84/3400)

=0.024908/0.072504

=0.34354*10^7

=3.4353 *10^6 fts.

1mi =5280 ft

=(3.4353/5280)*10^6

=650.644 miles

Now

For Endurance

E=(η/c)*{(Cl^3/2)/Cd}*(2*ρ*S)^1/2*[1/(W1)^1/2 -1/(W0)^1/2].

=(0.8/2.12*10^-7)*{(0.4110^3/2)/0.0342}*(2*0.002377*174)^1/2*[1/(3151.84)^1/2 -1/(3400)^1/2]

=3.7735*10^6*7.7043*0.8272*0.0006629

=0.01927*10^6

=1.927*10^4 sec

here 1hr =3600 sec

E=(1.927/3600)*10^4

=5.3528 hrs

7 0

2 years ago