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alekssr [168]
3 years ago
13

How can goal setting help with academic performance?

Engineering
1 answer:
Zina [86]3 years ago
8 0

Answer:

B.)It is an effective study skill.

Explanation:

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List and briefly describe two modern's materials needs
Angelina_Jolie [31]

Answer:

Modern and smart materials for making the products are improved by developing new materials and find new uses for the existing. As, modern industrialization society is increased demand and quality of the product.

Two modern's materials are:

Carbon Fiber: As, carbon fiber is a strong material and it is light in weight. Designers used it because it is five times strong as steel and two times as stiff. Carbon fiber is basically made out of very thin strands of carbon.

Fiber Optics: It is a new technology as, it is used as transparent solid to transmitted light signals.

4 0
3 years ago
Typically each development platform consists of the following components, except:Select one:a.Operating systemb.System softwarec
damaskus [11]

Typically each development platform consists of the following components except compilers and assemblers

  • The platform development simply means the development of the fundamental software which is vital in making hardware work.

  • Operating system: This refers to the low-level software that communicates with the hardware so that other programs can be able to run.

  • System software: This is the software that's designed in order to provide a platform for the other software. Examples include search engines, Microsoft Windows, etc.

  • Compilers and assemblers: Compliers are sued in converting source code to a machine-level language. Assembler is used in converting assembly code to machine code.

  • Hardware platform: This is a set of hardware where the software applications are run.

In conclusion, the correct option is Compilers and assemblers.

Read related link on:

brainly.com/question/21650058

4 0
2 years ago
Part A Engineering stress and strain are calculated using the actual cross-sectional area and length of the specimen. True or fa
Galina-37 [17]

Answer: True

Explanation:

Engineering stress is the applied load divided by the original cross-sectional area of a material. It is also known as nominal stress. It can also be defined as the force per unit area of a material. Engineering Stress is usually in large numbers.

While Engineering strain is the amount that a material deforms per unit length in a tensile test.  It can also be defined as extension per unit length. It has no unit as it is a ratio of lengths. Engineering Strain is in small numbers.

5 0
3 years ago
Read 2 more answers
can anyone give me tips on adding HP to my 2014 dodge charger SE? it uses the 3.6L220 CI 24 valve v6 vvt (variable valve timing)
marissa [1.9K]
<h2>ANSWER</h2><h2></h2>

I had a couple of answers for this, but when I checked nothing

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<h2></h2>

5 0
3 years ago
A hydraulic jump is induced in an 80 ft wide channel. The water depths on either side of the jump are 1 ft and 10 ft. Please cal
Sphinxa [80]

Answer:

a) The velocity is 42.0833 ft/s

b) The flow rate is 3366.664 ft³/s

c) The Froude number is 0.2345

d) The flow energy dissipated (expressed as percentage of the energy prior to the jump) is 18.225 ft

e) The critical depth is 3.8030 ft

Explanation:

Given data:

80 ft wide channel, L

1 ft and 10 ft water depths, d₁ and d₂

Questions: a) Velocity of the faster moving flow, v = ?

b) The flow rate (discharge), q = ?

c) The Froude number, F = ?

d) The flow energy dissipated, E = ?

e) The critical depth, dc = ?

a) For the velocity:

\frac{d_{2} }{d_{1} } =\frac{1}{2} (\sqrt{1+8F^{2} } -1)

10*2=\sqrt{1+8F^{2} } -1

Solving for F:

F = 7.4162

v=F\sqrt{gd_{1} }

Here, g = gravity = 32.2 ft/s²

v=7.4162*\sqrt{32.2*1} =42.0833ft/s

b) The flow rate:

q=v*L*d_{1} =42.0833*80*1=3366.664ft^{3} /s

c) The Froude number:

v_{2} =\frac{q}{L*d_{2} } =\frac{3366.664}{80*10} =4.2083ft/s

F=\frac{v_{2}}{\sqrt{gd_{2} } } =\frac{4.2083}{\sqrt{32.2*10} } =0.2345

d) The flow energy dissipated:

E=\frac{(d_{2}-d_{1})^{3} }{4d_{1}d_{2}} =\frac{(10-1)^{3} }{4*1*10} =18.225ft

e) The critical depth:

d_{c} =(\frac{(\frac{q}{L})^{2}  }{g} )^{1/3} =(\frac{(\frac{3366.664}{80})^{2}  }{32.2} )^{1/3} =3.8030ft

4 0
3 years ago
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