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2 years ago

9

Many people use microwave ovens to cook their food. Which option is best described as an engineering endeavor related to microwa

ve ovens? APEX

2 answers:

lyudmila [28]2 years ago

7 0

Answer:

researching the way people use microwave ovens to determine how to improve their design.

Explanation:

Apex. hope this helps !!!

vovangra [49]2 years ago

6 0

Answer:

the researching one

should be B or A

Explanation:

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Which of the following has special properties that allow forces and pressure to be distributed evenly?

Thepotemich [5.8K]

Answer:

Fluids

Explanation:

Fluids has special properties that allow forces and pressure to be distributed evenly within them.

Fluids are gases and liquids whose intermolecular forces of attraction are generally weak or non-existence.
Therefore, when pressure is applied to them, it permeates evenly on all parts.
Their ability to tend to randomness makes liquids and gases very viable for distributing pressure.

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3 years ago

Who is the inventor that created the light bulb

uranmaximum [27]

Answer:

Thomas Elva Edison

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People with skills and training in areas such as marketing or accounting are an important part of the manufacturing industry.

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Answer:

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2 years ago

A geothermal heat pump absorbs 15 KJ/s of heat from the Earth 15 m below a house. This heat pump uses a 7.45 kJ/s compressor.

Anna007 [38]

Answer:

COP of the heat pump is 3.013

OP of the cycle is 1.124

Explanation:

W = Q₂ - Q₁

Given

a)

Q₂ = Q₁ + W

= 15 + 7.45

= 22.45 kw

COP = Q₂ / W = 22.45 / 7.45 = 3.013

b)

Q₂ = 15 x 1.055 = 15.825 kw

therefore,

Q₁ = Q₂ - W

Q₁ = 15.825 - 7.45 = 8.375

∴ COP = Q₁ / W = 8.375 / 7.45 = 1.124

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3 years ago

Read 2 more answers

The fracture toughness of a stainless steel is 137 MPa*m12. What is the tensile impact load sustainable before fracture that a r

Charra [1.4K]

Answer:

7.7 kN

Explanation:

The capacity of a material having a crack to withstand fracture is referred to as fracture toughness.

It can be expressed by using the formula:

$K = \sigma Y \sqrt{\pi a}$

where;

fracture toughness K = 137 MPa $m^{1/2}$

geometry factor Y = 1

applied stress $\sigma$ = ???

crack length a = 2mm = 0.002

∴

$137 =\sigma \times 1 \sqrt{ \pi \times 0.002 }$

$137 =\sigma \times 0.07926$

$\dfrac{137}{0.07926} =\sigma$

$\sigma = 1728.489 MPa$

Now, the tensile impact obtained is:

$\sigma = \dfrac{P}{A}$

P = A × σ

P = 1728.289 × 4.5

P = 7777.30 N

P = 7.7 kN

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3 years ago