A beam of span L meters simply supported by the ends, carries a central load W. The beam section is shown in figure. If the maximum shear stress is 450 N/cm2 when the maximum bending stress is 1500 N/cm2. Calculate the value of the centrally applied point load W and the span L. The overall height of the I section is 29 cm.

Answer:

CLIMATE CHANGE HAS inexorably stacked the deck in favor of bigger and more intense fires across the American West over the past few decades, science has incontrovertibly shown. Increasing heat, changing rain and snow patterns, shifts in plant communities, and other climate-related changes have vastly increased the likelihood that fires will start more often and burn more intensely and widely than they have in the past.

Explanation:

Answer:

On July 13, 1787, Congress enacts the Northwest Ordinance, structuring settlement of the Northwest Territory and creating a policy for the addition of new states to the nation. ... In 1781, Virginia began by ceding its extensive land claims to Congress, a move that made other states more comfortable in doing the same

Answer:

They divided up the Northwest Territory into acre squares and sold them.

Explanation:

Answer:

mechanical engineer is the best answer

Answer:

1.5510 mm

Explanation:

Plane strain fracture toughness = 45 MPa√m

failing stress ( б ) = 300 MPa

maximum length of surface crack ( a )= 0.95 mm

Determine maximum allowable surface crack length ( in mm )

we will make use of this relationship for  Design stress equation to determine the value of Y

б = [tex]\frac{k}{y\sqrt{\pi *a} }[/tex]    --------- ( 1 )

k = 45 MPa√m

б  = 300 MPa

a = 0.95 mm

y = ?

From equation 1 make Y subject of the equation ( also substitute values into equation 1 above )

hence ; y = 2.7457

Now determine maximum allowable surface crack when component is exposed to a stress of 300 MPa and made from another alloy with plane-strain fracture toughness of 57.5 MPa√m

we will apply the equation

б = [tex]\frac{k}{y\sqrt{\pi *a} }[/tex]    --------- ( 2 )

K = 57.5 MPa√m

б = 300 MPa

y = 2.7457

a ( maximum allowable surface crack ) = ?

from equation make a subject of the equation

a = [tex]\frac{1}{\pi } (\frac{k}{\alpha y} )^{2}[/tex]  

a = [tex]\frac{1}{\pi } (\frac{57.5}{300*2.7457} )^2[/tex]  =   1.5510 mm

Answer:

Hello your question is incomplete  below is the complete question

Electronic components are often mounted with good heat conduction paths to a finned aluminum base plate, which is exposed to a stream of cooling air from a fan. The sum of the mass times specific heat products for a base plate and components is 5000 J/K, and the effective heat transfer coefficient times surface area product is 10 W/K. The initial temperature of the plate and the cooling air temperature are 295 K when 300 W of power are switched on. 1) Find the plate temperature after 10 minutes.

answer ; 311.36 k

Explanation:

Given data :

sum of mass * specific heat products for a base plate and components ( Mcp )

= 5000 J/K

effective heat transfer coefficient * surface area ( hA )  = 10 W/K

Initial temperature of plate and cooling air temperature( Tc ) = 295 k

power ( Q = W ) = 300 W

a) Determine plate temperature after 10 minutes

10 mins = 600 secs ( t )

heat supplied = change in temp + heat loss

          Q * t    = mCp ( ΔT ) + hA ( ΔT ) t

   300*600   =  5000 * ( T -295 )  + 10 ( T -295 ) * 600

therefore ; T - 295 = 16.363

                           T = 311.36 K

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

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:

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:

Less Friction Error

Explanation:

The friction error is very less in the moving

iron instrument because their torque weight

ratio is high.

hope it helps!

Answer:

Universal Use -The MI instrument is independent of the of current and hence for both AC and DC

Explanation:

Mark as Brainlist Answer

A power washer is being used to clean the siding of a house. Water at the rate of 0.1 kg/s enters at 20°c and 1 atm, with the velocity 0.2m/s. The jet of water exits at 23°c, 1 atm with a velocity 20m/s at an elevation of 5m. At steady state, the magnitude of the heat transfer rate from power unit to the surroundings is 10% of the power input. Determine the power input to the motor in kW.

Answer:

Net power of 1.2 KW is being extracted

Explanation:

We are given;

Mass flow rate; m' = 0.1kg/s

Inlet temperature; T1 = 20°C = 293K

Inlet pressure; P1 = 1 atm = 10^(5) pa

Inlet velocity; v1 = 0.2 m/s

Exit Pressure; P2 = 1 atm = 10^(5) pa

Exit Temperature; T2 = 1 atm = 296K

Exit velocity; V2 = 20m/s

Change in elevation; h = Z2 - Z1 = 5m

We are told that the magnitude of the heat transfer rate from the power unit to the surroundings is 10% of the power input.

Thus;

Q = -0.1W

From Bernoulli equation;

Q - W = ∆Potential energy + ∆Kinetic energy + ∆Pressure energy

Where;

∆Potential energy = mg(z2 - z1)

∆Kinetic energy = ½m(v2² - v1²)

∆Pressure energy = mc_p(T2 - T1)

Thus;

-0.1W - W = [m'g(z2 - z1)] + [½m'(v2² - v1²)] + [m'c_p(T2 - T1)]

Where C_p is specific heat capacity of water = 4200 J/Kg.k

Plugging in the relevant values, we have;

-1.1W = (0.1 × 9.81 × 5) + (½ × 0.1(20² - 0.2²)) + (0.1 × 4200 × (296 - 293))

-1.1W = 4.905 + 19.998 + 1260

-1.1W = 1284.903

W = -1284.903/1.1

W ≈ -1168 J/s ≈ -1.2 KW

The negative sign means that work is extracted from the system.

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]

Technicians A and B both are right with respect to their thinking and determination. Thus, the correct option is C.

Collision energy may be defined as a circumstance in which two or more bodies or particles come together with a resulting exchange of energy and alteration of direction.

Both pillar and floor pan reinforcement may be designed to transfer collision energy through the most prominent way to explain this mechanism of collisions.

A pillar exerts complete pressure on the floor in order to support the roof of the building, while floor pan reinforcement performs the same mechanism of collision energy with a different mechanism of action.

Therefore, technicians A and B both are right with respect to their thinking and determination. Thus, the correct option is C.

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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 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:

[tex]3.32\ \text{kW}[/tex]

Explanation:

[tex]T_c[/tex] = Outside temperature = [tex]-5^{\circ}\text{C}[/tex]

[tex]T_h[/tex] = Temperature of room = [tex]21^{\circ}\text{C}[/tex]

[tex]Q_h[/tex] = Heat loss = 135000 kJ/h = [tex]\dfrac{135000}{3600}=37.5\ \text{kW}[/tex]

Coefficient of performance of heat pump

[tex]\text{COP}=\dfrac{1}{1-\dfrac{T_c}{T_h}}\\\Rightarrow \text{COP}=\dfrac{1}{1-\dfrac{273.15-5}{273.15+21}}\\\Rightarrow \text{COP}=11.3[/tex]

Input power

[tex]W_i=\dfrac{Q_h}{\text{COP}}\\\Rightarrow W_i=\dfrac{37.5}{11.3}\\\Rightarrow W_i=3.32\ \text{kW}[/tex]

The minimum power required to drive this heat pump is [tex]3.32\ \text{kW}[/tex].

Answer:

false

Explanation:

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:

Maybe when there is a fire there can be fire drones that can take it out. and it can also resuce people who are stuck there.

Explanation:

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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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

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:

FOLLOWED BY...  2 6 5 3 5 8 9

UwU

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

Answer:

Air at 1 atm and 25◦C blows across a large concrete surface 20 m wide maintained

at 60◦C. The flow velocity is 6 m/s. Calculate the convection heat loss from the

surface.

This is heat transfer convection, mechanical engineering

please solve this question guys I'm gonna really really be appreciate it for you guys

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:

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        

Answer:

The ability to read electrical schematics is a really useful skill to have. To start developing your schematic reading abilities, it's important to memorize the most common schematic symbols. ... You should also be able to get a rough idea of how the circuit works, just by looking at the schematic.

Explanation: