All categories

2 years ago

A barometer reads a height of 78 cmHg. Express this atmospheric pressure to:

Engineering

2 answers:

Yakvenalex [24]2 years ago

7 0

A barometer reads a height of 78 cmHg. Express this atmospheric pressure to $\large\mathsf{\red{\underline{Pascal(pa)}}}$

balu736 [363]2 years ago

4 0

Answer:

Pascal (Pa)

hope it helps uh

You might be interested in

How many sets of equations (V and M equations) would you need to describe shear and moment as functions of x for this beam? In o

den301095 [7]

Shear and moment as functions of x is described below .

Explanation:

1. Beam is the slender bar that carries transverse

loading; that is, the applied force are perpendicular to the bar.

2. In a beam, the internal force system consist of a shear force and

a bending moment acting on the cross section of the bar.

3. The shear force and the bending moment usually vary continuously

along the length of the beam.

4. The internal forces give rise to two kinds of stresses on a

transverse section of a beam:

(1) normal stress that is caused by

bending moment and

(2) shear stress due to the shear force.

Knowing the distribution of the shear force and the bending

moment in a beam is essential for the computation of stresses

and deformations.

Shear- Moment Equations

The determination of the internal force system acting at a given

section of a beam : draw a free-body diagram that expose these

forces and then compute the forces using equilibrium equations.

The goal of the beam analysis －determine the shear force V and the bending moment M at every cross section of the beam.

To derive the expressions for V and M in terms of the distance x

measured along the beam. By plotting these expressions to scale,

obtain the shear force and bending moment diagrams for the

beam.

The shear force and bending moment diagrams are convenient

visual references to the internal forces in a beam; in particular, they identify the maximum values of V and M

5 0

2 years ago

Blank is a measure of the amount of matter in a given amount of space

madam [21]

Hope this helps you buddy!!!!! :}

Mass The measure of the amount of matter in an object.

5 0

2 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

3 years ago

A 1000-MVA, 20-kV, 60-Hz, three-phase generator is connected through a 1000-MVA, 20-kV, Dy345-kV, Y transformer to a 345-kV circ

aniked [119]

Answer:

(a) the subtransient current through the breaker in per-unit and in kA rms = 71316.39kA

(b) the rms asymmetrical fault current the breaker interrupts, assuming maximum dc offset. = 152KA

Explanation:

check the attached files for explanation

7 0

3 years ago

Very thin films are usually deposited under vacuum conditions to prevent contamination and ensure that atoms can fly directly fr

katrin [286]

Answer:

a. 9947 m

b. 99476 times

c. 2*10^11 molecules

Explanation:

a) To find the mean free path of the air molecules you use the following formula:

$\lambda=\frac{RT}{\sqrt{2}\pi d^2N_AP}$

R: ideal gas constant = 8.3144 Pam^3/mol K

P: pressure = 1.5*10^{-6} Pa

T: temperature = 300K

N_A: Avogadros' constant = 2.022*10^{23}molecules/mol

d: diameter of the particle = 0.25nm=0.25*10^-9m

By replacing all these values you obtain:

$\lambda=\frac{(8.3144 Pa m^3/mol K)(300K)}{\sqrt{2}\pi (0.25*10^{-9}m)^2(6.02*10^{23})(1.5*10^{-6}Pa)}=9947.62m$

b) If we assume that the molecule, at the average, is at the center of the chamber, the times the molecule will collide is:

$n_{collision}=\frac{9947.62m}{0.05m}\approx198952\ times$

c) By using the equation of the ideal gases you obtain:

$PV=NRT\\\\N=\frac{PV}{RT}=\frac{(1.5*10^{-6}Pa)(\frac{4}{3}\pi(0.05m)^3)}{(8.3144Pa\ m^3/mol\ K)(300K)}=3.14*10^{-13}mol\\\\n=(3.14*10^{-13})(6.02*10^{23})\ molecules\approx2*10^{11}\ molecules$

5 0

2 years ago