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Tanzania [10]
3 years ago
12

Question 9 of 25

Engineering
1 answer:
mafiozo [28]3 years ago
3 0

Answer:

D

Explanation:

took test failed question D is the right answer

You might be interested in
¿Qué aditivo se debe incorporar a la masa de hormigón para aumentar su resistencia frente a los ciclos alternados de hielo-deshi
tamaranim1 [39]

Answer:

Los aditivos que deben incorporarse a la masa de concreto para aumentar su resistencia a los ciclos alternos de congelación y descongelación son;

1. Agentes de arrastre de aire (AEA) o

2. Materiales poliméricos súper absorbentes

Explanation:

La resistencia alterna de los ciclos de congelación y descongelación en el concreto puede aumentarse mediante la adición de agentes de arrastre de aire.(AEA) que es un surfactante, crea burbujas de aire muy pequeñas en el concreto resultante para mejorar la durabilidad y resistencia del cemento al ciclo repetido de congelación y descongelación o materiales poliméricos súper absorbentes

Ejemplos de agentes de arrastre de aire son;

Sulfonatos alcalinos

Acidos de resinas sulfonadas

Sales de ácidos grasos

Ejemplos de materiales poliméricos superabsorbentes son;

SAP0.26CT

SAP0.39PT.

6 0
3 years ago
A rigid, well-insulated tank of volume 0.9 m is initially evacuated. At time t = 0, air from the surroundings at 1 bar, 27°C beg
Eva8 [605]

Answer:

\dot{w}= -0.303 KW

Explanation:

This is the case of unsteady flow process because properties are changing with time.

From first law of thermodynamics for unsteady flow process

\dfrac{dU}{dt}=\dot{m_i}h_i+\dot{Q}-\dot{m_e}h_i+\dot{w}

Given that tank is insulated so\dot{Q}=0 and no mass is leaving so

\dot{m_e}=0

\int dU=\int \dot{m_i}h_i\ dt-\int \dot{w}\ dt

m_2u_2-m_1u_1=(m_2-m_1)h_i- \dot{w}\Delta t

Mass conservation m_2-m_1=m_e-m_i

m_1,m_2 is the initial and final mass in the system respectively.

Initially tank is evacuated so m_1=0

We know that for air u=C_vT ,h=C_p T,P_2v_2=m_2RT_2

m_2=0.42 kg

So now putting values

0.42 \times 0.71 \times 730=0.42\times 1.005\times 300- \dot{w} \times 300

\dot{w}= -0.303 KW

3 0
3 years ago
technician a says that a gasoline engine is designed to operate at near or maximum speed for long periods without damage. techni
Vinil7 [7]

we can see that the correct person here will be Technician B who says that a diesel engine is designed to operate at near or maximum speed for long periods without damage.

<h3>What is a diesel engine?</h3>

A diesel engine is actually known to be an engine that makes use of diesel as its fuel. In other words, diesel engines run on diesel.

We see here that a diesel engine is actually designed to work for a very time on near or maximum speed without damage. This is true because the diesel fuel has that strength.

Also, diesel engines may be designed as two or four stroke cycles.

Learn more about diesel engine on brainly.com/question/13146091

#SPJ1

3 0
1 year ago
A westbound section of freeway currently has three 12-ft wide lanes, a 6-ft right shoulder, and no ramps within 3 miles upstream
Tresset [83]

Answer:

•Estimated density = 39.685Pc/mi/en

•Level of service, LOS frequency = LOSC

Explanation:

We are given:

•Freeway current lane width,B = 12ft

• freeway current shoulder width,b = 6ft

• percentage of heavy vehicle, Ptb = 10℅

• peak hour factor, PHF = 0.9

Let's consider,

•Number of lanes N = 4

• flow of traffic V = 7500vph

• percentage of Rv = 0, therefore the freeflow speed in freeway FFS = 70mph

• cars equivalent for recreational purpose Er= 2

•cars to be used for trucks and busses Etb= 2.5

Let's first calculate for the heavy adjustment factor.

We have:

F_H_v = \frac{1}{1+P_t_b(E_t_b-1)+Pr(Er-1)}

Substituting figures in the equation we have:

= \frac{1}{1+0.1(2.5-1)+0(2-1)}

= 0.75

Let's now calculate equivalent flow rate of the car using:

Vp = \frac{V}{(P_H_F)*N)*(F_H_v)*(F_p)}

= \frac{7500}{0.9*4*0.75*1}

= 2777.7 pc/h/en

Calculating for traffic density, we have:

D = \frac{Vp}{FFS}

D = \frac{2777.7}{70}

D = 39.685 Pc/mi/en

Using the table for LOS criteria of basic frequency segment, the level of service LOS of frequency is LOSC

4 0
3 years ago
What is the voltage output (in V) of a transformer used for rechargeable flashlight batteries, if its primary has 515 turns, its
kow [346]
<h2>Answer:</h2>

7532V

<h2>Explanation:</h2>

For a given transformer, the ratio of the number of turns in its primary coil (N_{p}) to the number of turns in its secondary coil (N_{s}) is equal to the ratio of the input voltage (V_{p}) to the output voltage (V_{s}) of the transformer. i.e

\frac{N_p}{N_s} = \frac{V_p}{V_s}            ----------------(i)

<em>From the question;</em>

N_{p} = number of turns in the primary coil = 8 turns

N_{s} = number of turns in the secondary coil = 515 turns

V_{p} = input voltage = 117V

<em>Substitute these values into equation (i) as follows;</em>

\frac{8}{515} = \frac{117}{V_s}

<em>Solve for </em>V_{s}<em>;</em>

V_{s} = 117 x 515 / 8

V_{s} = 7532V

Therefore, the output voltage (in V) of the transformer is 7532

6 0
3 years ago
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