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3 years ago

Please help i will give brainilest

Engineering

2 answers:

Eduardwww [97]3 years ago

8 0

Answer:

I hot u

Explanation:

it makes things function better and more fluently than a manual version of the certain object

Troyanec [42]3 years ago

3 0

Answer:

l m a o

Explanationl

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What is the difference between a single-model production line and a mixed-model production line?

Nat2105 [25]

The unique model production line is responsible for producing identical pieces. For this purpose the balancing of the assembly line is only responsible for assembling a model throughout the line.

This is a considerable difference compared to the mixed model assembly line where many models are assembled during the same production line, that is, it produces parts or products that have slight changes accommodated in them, with slight variations in their model or products of soft variety

The choice of the type of production depends on the type of company and its own demand, always prioritizing the efficiency in the operation. Generally, the mixed model tends to be chosen when demand is very large and customer demand is required to be met. In others it is considered a plant model in which half of the line is mixed and the other one is the only model in order to keep the efficiency balanced.

6 0

3 years ago

Determine the combined moment about O due to the weight of the mailbox and the cross member AB. The mailbox weighs 3.2 lb and th

koban [17]

Answer:

Attached is the complete question but the weight of the mailbox and cross bar differs from the given values which are : weight of mail box = 3.2 Ib, weight of the uniform cross member = 10.3 Ib

Answer : moment of inertia = 186.7 Ib - in

Explanation:

Given data

weight of the mailbox = 3.2 Ib

weight of the uniform cross member = 10.3 Ib

The origin is of mailbox and cross member is 0

The perpendicular distance from Y axis of centroid of the mailbox

= 4 + (25/2) = 16.5"

The centroid of the bar =( ( 1 + 25 + 4 + 4 ) / 2 ) - 4 = 13"

therefore The moment of Inertia( Mo) = (3.2 * 16.5) + ( 10.3 * 13)

= 52.8 + 133.9 = 186.7 Ib-in

8 0

3 years ago

During a load test, a battery's voltage drops below a specific value. what action should the technician take?

Katyanochek1 [597]

Answer:

Explanation:

allow battery to become fully discharged and retest

4 0

2 years ago

Identify the advantages of using 6 tube passes instead of just 2 of the same diameter in shell-and-tube heat exchanger.What are

liq [111]

Answer:

Please check explanation for answer

Explanation:

Here, we are concerned with stating the advantages and disadvantages of using a 6 tube passes instead of a 2 tube passes of the same diameter:

<u>Advantages</u>

* By using a 6 tube passes diameter, we are increasing the surface area of the heat transfer surface

* As a result of increasing the heat transfer surface area, the rate of heat transfer automatically increases too

Thus, from the above, we can conclude that the heat transfer rate of a 6 tube passes is higher than that of a 2 tube passes of the same diameter.

<u>Disadvantages</u>

* They are larger in size and in weight when compared to a 2 tube passes of the same diameter and therefore does not find use in applications where space conservation is quite necessary.

* They are more expensive than the 2 tube passes of the same diameter and thus are primarily undesirable in terms of manufacturing costs

5 0

3 years ago

Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occu

Sonja [21]

Answer:

The rate of work developed is $\frac{\r W}{\r m}= 300kJ/kg$

The rate of entropy produced within the turbine is $\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

Explanation:

From the question we are told

The rate at which heat is transferred is $\frac{\r Q}{\r m } = - 30KJ/kg$

the negative sign because the heat is transferred from the turbine

The specific heat capacity of air is $c_p = 1.1KJ/kg \cdot K$

The inlet temperature is $T_1 = 970K$

The outlet temperature is $T_2 = 670K$

The pressure at the inlet of the turbine is $p_1 = 400 kPa$

The pressure at the exist of the turbine is $p_2 = 100kPa$

The temperature at outer surface is $T_s = 315K$

The individual gas constant of air R with a constant value $R = 0.287kJ/kg \cdot K$

The general equation for the turbine operating at steady state is \

$\r Q - \r W + \r m (h_1 - h_2) = 0$

h is the enthalpy of the turbine and it is mathematically represented as

$h = c_p T$

The above equation becomes

$\r Q - \r W + \r m c_p(T_1 - T_2) = 0$

$\frac{\r W}{\r m} = \frac{\r Q}{\r m} + c_p (T_1 -T_2)$

Where $\r Q$ is the heat transfer from the turbine

$\r W$ is the work output from the turbine

$\r m$ is the mass flow rate of air

$\frac{\r W}{\r m}$ is the rate of work developed

Substituting values

$\frac{\r W}{\r m} = (-30)+1.1(970-670)$

$\frac{\r W}{\r m}= 300kJ/kg$

The general balance equation for an entropy rate is represented mathematically as

$\frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma = 0$

=> $\frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)$

generally $(s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ]$

substituting for $(s_1 -s_2)$

$\frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} + c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ]$

Where $\frac{\sigma}{\r m}$ is the rate of entropy produced within the turbine

substituting values

$\frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ]$

$\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

5 0

3 years ago