The section should span between 10.9 and 13.4 cm (4.30 and 5.30 inches) as measured from the end supports and should be able to carry a minimum load of 13 N applied evenly to the top of the span. The maximum load for each member is 4.5 Newtons in compression and tension. Each member is 4.2 cm long (ball center to ball center). All joints must be fully constrained. Determine the max value of P before the truss fails. The section that can hold the most weight will receive bonus points.

Answer:

a) for back pressures of (a) 500 kPa

- mass flow rate is 2.127 kg/s

- exit Mach number is 1.046

b) for back pressures of (a) 784 kPa

- mass flow rate is 1.793 kg/s

- exit Mach number is 0.6

Explanation:

Given that;

A₂ = 0.001 m²

P₁ = 1 MPa

T₁ = 360 K

k = 1.4

P₂ = 500 Kpa

(1000/500)^(1.4-1 / 1.4) = 360 /T₂

2^(0.4/1.4) = 360/T₂

1.219 = 360 / T₂

T₂ = 360 / 1.219

T₂ = 295.32 K

CpT₁ + V₁²/2000 = CpT₂ + V₂²/2000

we substitute

CpT₁ + V₁²/2000 = CpT₂ + V₂²/2000

1.005 × 360 =  1.005 × 295.32 + v₂²/2000

v₂ = 360.56 m/s²

p₂v₂ = mRT₂

500 × (0.001 × 360.56) = m × 0.287 × 295.32

m = 2.127 kg/s

so Mach Number = V₂ / Vc

Vc = √( kRT) = √( 1.4 × 287 × 295.32) = 344.47 m/s

So Mach Number =  V₂ / Vc  =  360.56 / 344.47 = 1.046

Therefore for back pressures of (a) 500 kPa

- mass flow rate is 2.127 kg/s

- exit Mach number is 1.046

b)

AT P₂ = 784 kPa

(1000/784)^(1.4-1 / 1.4) = 360/T₂

T₂ = 335.82 K

now

V₂²/2000 = 1.005( 360 - 335.82)

V₂ = 220.45 m/s

P₂V₂ = mRT₂

784 × (0.001 × 220.45) = m( 0.287) ( 335.82)

172.83 = 96.38 m

m = 172.83 / 96.38

m = 1.793 kg/s

just like in a)

Vc = √( kRT) = √( 1.4 × 287 × 335.82) = 367.32 m/s

Mach Number = V₂ / Vc = 220.45 / 367.32 = 0.6

Therefore for back pressures of (a) 784 kPa

- mass flow rate is 1.793 kg/s

- exit Mach number is 0.6

Following are the

Given:

[tex]A_2 = 0.001\ m^2\\\\P_1 = 1\ MPa\\\\T_1 = 360\ K\\\\k = 1.4\\\\P_2 = 500\ Kpa\\\\[/tex]

To find:

Flow rate of mass, and Mach number

Solution:

For point a)

Using formula:

[tex]\to P^{\frac{r-1}{r}} \alpha T\\\\\to (\frac{1000}{500})^(\frac{1.4-1}{1.4}) = \frac{360}{T_2}\\\\\to 2^(\frac{0.4}{1.4}) = \frac{360}{T_2}\\\\\to 1.219 = \frac{360}{T_2}\\\\\to T_2 = \frac{360}{1.219}\\\\\to T_2 = 295.32\ K\\\\[/tex]

[tex]\to C_pT_1 + \frac{V_1^2}{2000} = C_pT_2 + \frac{V_2^2}{2000}\\\\\to 1.005 \times 360 = 1.005 \times 295.32 + \frac{v_2^2}{2000}\\\\\to v_2 = 360.56 \ \frac{m}{s^2} \\\\\to p_2v_2 = mRT_2\\\\\to 500 \times (0.001 \times 360.56) = m \times 0.287 \times 295.32\\\\\to m = 2.127\ \frac{kg}{s}\\\\[/tex]

Mach Number [tex]= \frac{V_2}{V_c}\\\\[/tex]

[tex]\to V_c = \sqrt{( kRT)} = \sqrt{( 1.4 \times 287 \times 295.32)} = 344.47 \ \frac{m}{s}\\\\[/tex]

[tex]\to \frac{V_2}{V_c} = \frac{360.56}{344.47} = 1.046[/tex]

For point b)

[tex]\to P_2 = 784\ kPa\\\\\to (\frac{1000}{784})^{(\frac{0.4}{1.4})} = \frac{360}{T_2}\\\\\to T_2 = 335.82\ K\\\\[/tex]

now

[tex]\to \frac{V_2^2}{2000} = 1.005( 360 - 335.82)\\\\\to V_2 = 220.45 \frac{m}{s}\\\\\to V_c = \sqrt{( kRT)} = \sqrt{( 1.4 \times 287 \times 335.82)} = 367.32\ \frac{ m}{s}\\\\\to c= \frac{220.45}{ 367.32} = 0.6\\[/tex]

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Answer: Rc = 400 Ω and Rb = 57.2 kΩ

Explanation:

Given that;

VCE = 5V

VCC = 15 V

iC = 25 mA

β = 100

VD₀ = 0.7 V

taking a look at the image; at loop 1

-VCC + (i × Rc) + VCE = 0

we substitute

-15 + ( 25 × Rc) + 5 = 0

25Rc = 10

Rc = 10 / 25

Rc = 0.4 k

Rc = 0.4 × 1000

Rc = 400 Ω

iC = βib

25mA = 100(ib)

ib = 25 mA / 100

ib = 0.25 mA

ib = 0.25 × 1000

ib = 250 μAmp

Now at Loop 2

-Vcc + (ib×Rb) + VD₀ = 0

-15 (250 × Rb) + 0.7 = 0

250Rb = 15 - 0.7

250Rb = 14.3

Rb = 14.3 / 250

Rb = 0.0572 μ

Rb = 0.0572 × 1000

Rb = 57.2 kΩ

Therefore Rc = 400 Ω and Rb = 57.2 kΩ

Answer:

lol

Explanation:r

You've asked an Incomplete question, lacking options. I answered based on the existing O*NET report.

Answer:

high school diploma

Explanation:

According to the Occupational Information Network (O*NET), most people who are Licensing Examiners and Inspectors typically have a high school diploma.

In other words, they do not seek to acquire a post-secondary school education.

Answer:

B

Explanation:

According to edge its answer B

associate's degree or on-the-job experience

got it right as a lucky guess as the O*net site is updated but edge doesn't bother to update their questions or links.

Click this link to view O*NET’s Education section for Licensing Examiners and Inspectors. According to O*NET, what is the most common level of education among Licensing Examiners and Inspectors?  

bachelor’s degree

associate's degree or on-the-job experience .......This is the correct answer.

some college, no degree

associate degree

70.40mol cuz

1g sodium sulfate = 0.00704mol

take 10kg × 1000 = 10,000g

10,000g × 0.00704

final answer 70.40mol

(as per my thinking)

Answer:

70.40mol cuz

1g sodium sulfate = 0.00704mol

take 10kg × 1000 = 10,000g

10,000g × 0.00704

final answer 70.40mol

Answer:

Para x=0:

[tex]\phi=1.2 W/m^{2}[/tex]  

Para x=30 cm:

[tex]\phi=-2.4 W/m^{2}[/tex]  

Explanation

Podemos utilizar la ley de Fourier par determinar el flujo de calor:

[tex]\phi=-k\frac{dT}{dx}[/tex](1)

Por lo tanto debemos encontrar la derivada de T(x) con respecto a x primero.

Usando la ley de potencia para la derivda, tenemos:

[tex]\frac{dT(x)}{dx}=300x-30[/tex]

Remplezando esta derivada en (1):

[tex]\phi=-0.04(300x-30)[/tex]

Para x=0:

[tex]\phi=0.04(30)[/tex]

[tex]\phi=1.2 W/m^{2}[/tex]  

Para x=30 cm:

[tex]\phi=-0.04(300*0.3-30)[/tex]

[tex]\phi=-2.4 W/m^{2}[/tex]    

Espero que te haya ayudado!

Answer:

When are resistors in series? Resistors are in series whenever the flow of charge, called the current, must flow through devices sequentially. For example, if current flows through a person holding a screwdriver and into the Earth, then  

R

1

 in Figure 1(a) could be the resistance of the screwdriver’s shaft,  

R

2

 the resistance of its handle,  

R

3

 the person’s body resistance, and  

R

4

 the resistance of her shoes.

Figure 2 shows resistors in series connected to a voltage source. It seems reasonable that the total resistance is the sum of the individual resistances, considering that the current has to pass through each resistor in sequence. (This fact would be an advantage to a person wishing to avoid an electrical shock, who could reduce the current by wearing high-resistance rubber-soled shoes. It could be a disadvantage if one of the resistances were a faulty high-resistance cord to an appliance that would reduce the operating current.)

Two electrical circuits are compared. The first one has three resistors, R sub one, R sub two, and R sub three, connected in series with a voltage source V to form a closed circuit. The first circuit is equivalent to the second circuit, which has a single resistor R sub s connected to a voltage source V. Both circuits carry a current I, which starts from the positive end of the voltage source and moves in a clockwise direction around the circuit.

Figure 2. Three resistors connected in series to a battery (left) and the equivalent single or series resistance (right).

To verify that resistances in series do indeed add, let us consider the loss of electrical power, called a voltage drop, in each resistor in Figure 2.

According to Ohm’s law, the voltage drop,  

V

, across a resistor when a current flows through it is calculated using the equation  

V

=

I

R

, where  

I

 equals the current in amps (A) and  

R

 is the resistance in ohms  

(

Ω

)

. Another way to think of this is that  

V

 is the voltage necessary to make a current  

I

 flow through a resistance  

R

.

So the voltage drop across  

R

1

 is  

V

1

=

I

R

1

, that across  

R

2

 is  

V

2

=

I

R

2

, and that across  

R

3

 is  

V

3

=

I

R

3

. The sum of these voltages equals the voltage output of the source; that is,

V

=

V

1

+

V

2

+

V

3

.

This equation is based on the conservation of energy and conservation of charge. Electrical potential energy can be described by the equation  

P

E

=

q

V

, where  

q

 is the electric charge and  

V

 is the voltage. Thus the energy supplied by the source is  

q

V

, while that dissipated by the resistors is

q

V

1

+

q

V

2

+

q

V

3

.

Explanation:

Connecting rods undergo a process to alleviate manufacturing stresses from forging, a process known as cold forging.​

Forging is a metalworking process that involves shaping metal with localized compressive forces. A hammer or a die is used to deliver the blows.

Forging is frequently classified according to temperature: cold forging, warm forging, or hot forging.

The goal of forging is to make metal parts. Metal forging produces some of the most durable manufactured parts available when compared to other manufacturing methods.

Minor cracks and empty spaces in the metal are filled as the metal is heated and pressed.

Cold forging is the process of deforming a metal material at room temperature using extremely high pressure. The slug is placed in a die and compressed by a press until it fits into the desired shape.

Thus, the answer is cold forging.

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Answer:

Without the indispensable advice of the business community, TVET Institutions will be unable to cover the gap in career knowledge required by the business community.  To develop workers who possess the knowledge and skills required by today's business entities, there is always the continual need for the educational institutions (gown) to regularly meet the business community (town).  This meeting provides the necessary ground for the institutions to develop programs that groom the workforce with skills that are needed in the current workplace.  Educational institutions that do not seek this important advice from the business community risk developing workers with outdated skills.

Explanation:

TVET Institutions mean Technical and Vocational Education and Training Institutions.  They play an important role in equipping young people to enter the world of work.  They also continue to develop programs that will improve the employability of workers throughout their careers.  They regularly respond to the changing labor market needs, adopt new training strategies and technologies, and expand the outreach of their training to current workers while grooming the young people for work.

Answer:

7.47 lb. s/ft

22.42  lb. s/ft

Explanation:

Without mincing words let us dive straight into the solution to the above problem.

STEP ONE: calculate or determine the frequency.

The frequency can be calculated by making use of the formula given below;

frequency, w = √k/m. Thus, frequency = √ 12 × 15/ [40/32.2] = 12 rad/s.

STEP TWO: Determine or calculate for the viscous damping coefficient, c  for damping ratio of 0.5 and 1.5 respectively.

The viscous damping coefficient, c for 0.5 = [ 12 × 0.5 × 2 ×[ 40/32.2] / 2= 7.47 lb. s/ft.

The viscous damping coefficient, c for 1.5= [ 12 × 1.5 × 2 ×[ 40/32.2] / 2 = 22.42  lb. s/ft.

Answer:

D, meter.

Explanation:

Rhythm is associated with meter, which identifies units of stressed and unstressed syllables.

If a poem has a regular rhythm throughout the poem, it has a meter. Option D is correct.

Meter, which distinguishes between stressed and unstressed syllables, is related to rhythm. The fundamental rhythmic framework of a stanza or a line of poetry is known as meter.

The number of feet in the poem serves as a measure of the poem's meter, which is the rhythm of the language.

Many traditional poem forms call for a certain verse meter or a group of meters that alternate in a specified pattern. Prosody refers to both the study of meters and other types of versification, as well as their practical application.

Therefore, option D is correct.

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Answer:

The annualized cost is:

$299,272.

Explanation:

a) Data and Calculations:

Year 0  Capital cost of road construction = $6 million

Year 10 Major improvements cost = $17 million

Year 25 Replacement and upgrade cost = $29 million

Year 40 Federal government one-time tax credit = $12 million

Period of project analysis = 50 years

Cost of capital (discount rate) = 10%

Annualized cost at present value costs:

Amount spent  Discount Factor     Present value

$6 million           1                               $6,000,000

$17 million         0.386                          6,562,000

$29 million       0.092                          2,668,000

($12 million)      0.0222                         (266,400)

Total cost                                          $14,963,600

Annualized cost = $14,963,600/50 =  $299,272

b) The annualized cost for the road construction project, which is the annualized value of the net present costs of $14,963,600, is divided by 50.  Before obtaining the net present costs, the cash outflows, including the tax credit, are discounted to their present values, using the discount rate of 10%.  And then, the average of the cost is obtained by dividing the total net present cost into 50 years.

Answer:

hello your question is incomplete attached below is the complete question

answer : attached below

Explanation:

let ; x(t)  be a real value signal for x ( jw ) = 0 , |w| > 200[tex]\pi[/tex]

g(t) = x ( t ) sin ( 2000 [tex]\pi t )[/tex]

[tex]x_{1} (t) = \frac{1}{2} x(t) sin ( 4000\pi t )[/tex]

next we apply Fourier transform

attached below is the remaining part of the solution

Answer:

no

Explanation:

How many trees would it take to reverse climate change?

1.2 trillion trees

Crowther says planting 1.2 trillion trees would give a reduction "way above" that figure. To put that in context, global CO2 emissions are around 37 billion tons per year.Apr 17, 2019

Answer:

generalization

Explanation:

Please mark me brainliest I need to level up

Answer:

to increase tensile strength.

Explanation:

Steel as part of the materials used in the construction industries, or other manufacturing industries could be defective if it was not appropriately manufactured. This defects could be due to foreign materials in it, scratches, blisters etc which would lead to the tensile strength being reduced.

In-order to overcome this, there is need to carry out a technique called lamination. This technique is the process of manufacturing the steel in multiple layers so that the composite materials improves the strength of the steel and its stability. This helps to prevent its failure when used in construction.

Answer:

u do the same thing as part B but only add 100 k, I think, cuz I'm still in middle school but I mean if u see it asks u to do the same thing as B but C says that instead, u do it at half pressure and 100 k is higher temp so what its asking is to repeat b but the twist is u do it at half pressure and 100 k is the higher temp

hope this helps :)  

Answer:

Sound waves are compression waves that cause energy transfer in air molecules

Sound waves have to have a medium to travel through

Loud sounds have high amplitude and vibrate with more energy than soft sounds

Explanation:

Sound waves is a form of energy composed of compression and rare factions. Sound waves are compression waves that cause energy transfer in air molecules.

Sound is an example of a mechanical wave hence it requires a material medium for propagation.

The amplitude of a sound wave determines its loudness or volume. A larger amplitude implies that we will have a louder sound, and a smaller amplitude means that we will have a softer sound.

Answer:a

Ieieksdjd snsnsnsnsksks

Answer:

The answer is "Both A and B" are right

Explanation:

During the previous twenty years car producers have made significant advances in planning vehicle structures that give more noteworthy tenant insurance in planar accidents (Lund and Nolan 2003). Be that as it may, there has been little agreement with respect to the significance of rooftop strength in rollover crashes, just as the best strategy for surveying that strength. In 2006 one-fourth of lethally harmed traveler vehicle tenants were associated with crashes where vehicle rollover was considered the most hurtful occasion (Protection Establishment for Expressway Wellbeing, 2007). Numerous lethally harmed tenants in rollovers are unbelted, and some are totally or mostly launched out from the vehicle (Deutermann 2002).

There is difference concerning how underlying changes could influence launch hazard or the danger of injury for inhabitants who stay in the vehicle, paying little mind to belt use.

Answer:A Only

Explanation:

I Took the test

Answer:

A turbo blade has the best features from both other types of blade. The continuous, serrated edge makes for fast cutting while remaining smooth and clean. They are mostly used to cut a variety of materials, such as tile, stone, marble, granite, masonry, and many other building materials.

Explanation:

Answer:

Masonry Saw

Explanation:

Masonry Saws can be used to cut brick, ceramic, tile and or stone.

Answer:

joule

Explanation:

heat is express in joules in x Ray equipment

Answer:

the actual specific work required by the compressor to operate is 2425.88 kJ/kg

Explanation:

Given that;

h₁ = 1350 kJ/kg

h₂₅ = 3412 kJ/kg

compressor efficiency П_ise = 0.85

we know that;

compressor efficiency П_ise = isentropic work / actual work

П_ise = (h₂₅ - h₁) / (h₂ - h₁ )

so

0.85 =  (h₂₅ - h₁) / (actual work )

Actual work = (h₂₅ - h₁) / 0.85

Actual work = (3412 - 1350) / 0.85

Actual work = 2062 / 0.85

Actual work = 2425.88 kJ/kg

Therefore the actual specific work required by the compressor to operate is 2425.88 kJ/kg

Answer:

W = 11,416.6879 N

L ≈ 64.417 cm

Explanation:

The maximum shear stress, [tex]\tau_{max}[/tex], is given by the following formula;

[tex]\tau_{max} = \dfrac{W}{8 \cdot I_c \cdot t_w} \times \left (b\cdot h^2 - b\cdot h_w^2 + t_w \cdot h^2_w \right )[/tex]

[tex]t_w[/tex] = 1 cm = 0.01

h = 29 cm = 0.29 m

[tex]h_w[/tex] = 25 cm = 0.25 m

b = 15 cm = 0.15 m

[tex]I_c[/tex] = The centroidal moment of inertia

[tex]I_c = \dfrac{1}{12} \cdot \left (b \cdot h^3 - b \cdot h_w^3 + t_w \cdot h_w^3 \right )[/tex]

[tex]I_c[/tex] = 1/12*(0.15*0.29^3 - 0.15*0.25^3 + 0.01*0.25^3) = 1.2257083 × 10⁻⁴ m⁴

Substituting the known values gives;

[tex]I_c = \dfrac{1}{12} \cdot \left (0.15 \times 0.29^3 - 0.15 \times 0.25^3 + 0.01 \times 0.25^3 \right ) = 1.2257083\bar 3 \times 10^{-4}[/tex]

[tex]I_c[/tex] = 1.2257083[tex]\bar 3[/tex] × 10⁻⁴ m⁴

From which we have;

[tex]4,500,000 = \dfrac{W}{8 \times 1.225708\bar 3 \times 10 ^{-4}\times 0.01} \times \left (0.15 \times 0.29^2 - 0.15 \times 0.25^2 + 0.01 \times 0.25^2 \right )[/tex]

Which gives;

W = 11,416.6879 N

[tex]\sigma _{b.max} = \dfrac{M_c}{I_c}[/tex]

[tex]\sigma _{b.max}[/tex] = 1500 N/cm² = 15,000,000 N/m²

[tex]M_c[/tex] = 15,000,000 × 1.2257083 × 10⁻⁴ ≈ 1838.56245 N·m²

From Which we have;

[tex]M_{max} = \dfrac{W \cdot L}{4}[/tex]

[tex]L = \dfrac{4 \cdot M_{max}}{W} = \dfrac{4 \times 1838.5625}{11,416.6879} \approx 0.64417[/tex]

L ≈ 0.64417 m ≈ 64.417 cm.

Answer:

Q = [ mCp ( ΔT) ] [tex]_{cooling water }[/tex]

(ΔT)[tex]_{cooling water}[/tex] and  Q  is given

[tex]m_{cooling water}[/tex]  = [tex]\frac{Q}{Cp[ T_{out} - T_{in} ] }[/tex]

next the rate of condensation of the steam

Q = [ m[tex]h_{fg}[/tex] ][tex]_{steam}[/tex]

  [tex]m_{steam} = \frac{Q}{h_{fg} }[/tex]

Total resistance of the condenser is

R = [tex]\frac{Q}{change in T_{cooling water } }[/tex]

Explanation:

How will the rate of condensation of the steam and the mass flow rate of the cooling water can be determined

Q = [ mCp ( ΔT) ] [tex]_{cooling water }[/tex]

(ΔT)[tex]_{cooling water}[/tex] and  Q  is given

[tex]m_{cooling water}[/tex]  = [tex]\frac{Q}{Cp[ T_{out} - T_{in} ] }[/tex]

next the rate of condensation of the steam

Q = [ m[tex]h_{fg}[/tex] ][tex]_{steam}[/tex]

  [tex]m_{steam} = \frac{Q}{h_{fg} }[/tex]

Total resistance of the condenser is

R = [tex]\frac{Q}{change in T_{cooling water } }[/tex]

Given :

Two children are playing on a seesaw. The child on the left weighs 50 lbs. And the child on the right weighs 100 lbs.

To Find :

If the child on the left is 3 feet from the pivot, how far must the child on the right be from the pivot from the seesaw to be balanced.

Solution :

We know, for seesaw to balance :

[tex]m_1gd_1=m_2gd_2[/tex]

Here, [tex]d_1\ and \ d_2[/tex] is distance from origin from pivot point.

Putting all the values in the equation, we get :

[tex]50\times g\times 3=100\times g\times d_2\\\\d_2=1.5\ feet[/tex]

Therefore, distance of right child from pivot to balance the seesaw is 1.5 feet.

Hence, this is the required solution.

Answer:

Explanation:

The cylinder head sits on the engine and closes off the combustion chamber. The gap that remains between the cylinder head and the engine is completed by the head gasket. Another task of the cylinder head is to ensure the constant lubrication of the cylinder