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kolbaska11 [484]

2 years ago

7

The only way to know if a design will work in real-world conditions is to build a model, or prototype, based on the plan. This i

s part of which step in the Design Process?
a
Plan
b
Brainstorm
c
Create
d
Improve Design

2 answers:

satela [25.4K]2 years ago

7 0

Answer:

Brainstorm

Explanation:

Engineers use the engineering design process when brainstorming solutions to real-life problems; they develop these solutions by testing and redesigning prototypes that work within given constraints

Lady bird [3.3K]2 years ago

3 0

Answer:

redesign

Explanation:

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The resultant force is directed along the positive x axis and has a magnitude of 1330 N.

Andrew [12]

Answer:

the magnitude of F_A is 752 N

the direction theta of F_A is 57.9°

Explanations:

Given that,

Resultant force = 1330 N in x direction

∑Fx = R

from the diagram of the question which i uploaded along with this answer

FB = 800 N

FAsin∅ + FBcos30 = 1330 N

FAsin∅ = 1330 - (800 × cos30)

FA = 637.18 / sin∅

Now ∑Fx = 0

FAcos∅ - FBsin30 = 0

we substitute for FA

(637.18 / sin∅)cos∅ = 800 × sin30

637.18 / 800 × sin30 = sin∅/cos∅

and we know that { sin∅/cos∅ = tan∅)

so tan∅ = 1.59295

∅ = 57.88° ≈ 57.9°

THEREFORE FROM THE EQUATION

FA = 637.18 / sin∅

we substitute ∅

so FA = 637.18 / sin57.88

FA = 752 N

3 0

3 years ago

. A Carnot heat pump is to be used to heat a house and maintain it at 22 °C in winter. When the outdoor temperature remains at 3

max2010maxim [7]

Answer:

The Carnot heat pump must work 3.624 hours per day to keep the temperature constant inside the house.

Explanation:

The net heat daily loss of the house is:

$Q_{losses} = \left(76000\,\frac{kJ}{h}\right)\cdot (24\,h)$

$Q_{losses} = 1.824\times 10^{6}\,kJ$

In order to keep the house warm, given heat must be equal to heat losses:

$Q_{H} = Q_{losses}$

Besides, the Coefficient of Performance for a Carnot heat pump is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

Where,

$T_{L}$ - Temperature of the cold reservoir (Outdoors), measured in Kelvin.

$T_{H}$ - Temperature of the hot reservoir (House), measured in Kelvin.

Given that $T_{L} = 276.15\,K$ and $T_{H} = 295.15\,K$ , the Coefficient of Performance is:

$COP_{HP} = \frac{295.15\,K}{295.15\,K-276.15\,K}$

$COP_{HP} = 15.534$

For a real heat machine, the Coefficient of Performance is determined by the following expression:

$COP_{HP} = \frac{Q_{H}}{W}$

Where:

$Q_{H}$ - Heat received by the house, measured in kilojoules.

$W$ - Work consumed by the Carnot heat pump, measured in kilojoules.

The daily work consumed is now cleared in the previous expression:

$W = \frac{Q_{H}}{COP_{HP}}$

$W = \frac{1.824\times 10^{6}\,kJ}{15.534}$

$W = 117419.853\,kJ$

The working time is calculated by dividing this result by input power. That is:

$\Delta t = \frac{W}{\dot W}$

$\Delta t = \left(\frac{117419.853\,kJ}{9\,kW} \right)\cdot \left(\frac{1}{3600}\,\frac{h}{s}\right)$

$\Delta t = 3.624\,h$

The Carnot heat pump must work 3.624 hours per day to keep the temperature constant inside the house.

8 0

3 years ago

A water jet that leaves a nozzle at 60 m/s at a certain flow rate generating power of 250 kW by striking the buckets located on

wlad13 [49]

Answer:

The mass flow rate of jet =69.44 kg/s

Explanation:

Given that

velocity of jet v= 60 m/s

Power P=250 KW

As we know that force offered by water F

$F=\rho\ A \ v^2$

Power P= F.v

So now power given as

$P=\rho\ A \ v^3$

We know that mass flow rate = ρAv

$P=mass\ flow\ rate\ \times v^2$

250 x 1000 = mass flow rate x 3600

mass flow rate = 69.44 kg/s

So the mass flow rate of jet =69.44 kg/s

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3 years ago

in verification of ohms law the mass is 100g, initial length is 31, final length is 31.3 what is the extension?

MissTica

Answer:

24

Explanation:

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3 years ago

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Answer:

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Explanation:

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4 0

3 years ago

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