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stellarik [79]
3 years ago
6

A force is a push or pull in A.a circle B.an arc C.a straight line

Engineering
1 answer:
Bingel [31]3 years ago
4 0

Answer:

I think it is B: an arc

Explanation:

hope this helps mark as brainiest

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Two vertical, parallel clean glass plates are spaced a distance of 2mm apart. if the plates are placed in water, how high will t
Ulleksa [173]

Answer with Explanation:

The capillary rise in 2 parallel plates immersed in a liquid is given by the formula

h=\frac{2\sigma cos(\alpha )}{\rho gd}

where

\sigma is the surface tension of the liquid

\alpha is the contact angle of the liquid

\rho is density of liquid

'g' is acceleratioj due to gravity

'd' is seperation between thje plates

Part a) When the liquid is water:

For water and glass we have

\sigma =7.28\times 10^{-2}N/m

\alpha =0

\rho _{w}=1000kg/m^3

Applying the values we get

h=\frac{2\times 7.28\times 10^{-2}cos(0)}{1000\times 9.81\times 2\times 10^{-3}}=7.39mm

Part b) When the liquid is mercury:

For mercury and glass we have

\sigma =485.5\times 10^{-3}N/m

\alpha =138^o

\rho _{w}=13.6\times 10^{3}kg/m^3

Applying the values we get

h=\frac{2\times 485.5\times 10^{-3}cos(138)}{13.6\times 1000\times 9.81\times 2\times 10^{-3}}=-2.704mm

The negative sign indicates that there is depression in mercury in the tube.

4 0
3 years ago
What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 5 × 10-4
White raven [17]

Answer:

The question is a problem that requires the principles of fracture mechanics.

and we will need this equation below to get the Max. Stress that exist at the tip of an internal crack.

Explanation:

Max Stress, σ = 2σ₀√(α/ρ)

where,

σ₀  = Tensile stress = 190MPa = 1.9x10⁸Pa

α  = Length of the cracked surface = (4.5x10⁻²mm)/2 = 2.25x10⁻⁵m

ρ  = Radius of curvature of the cracked surface = 5x10⁻⁴mm = 5x10⁻⁷m

Max Stress, σ = 2 x  1.9x10⁸ x (2.25x10⁻⁵/5x10⁻⁷)⁰°⁵

Max Stress, σ = 2 x  1.9x10⁸ x 6.708 Pa

Max Stress, σ = 2549MPa

Hence, the magnitude of the maximum stress that exists at the tip of an internal crack = 2549MPa

7 0
4 years ago
Which type of Artificial Intelligence (AI) can repeatedly perform tasks of limited scope?
seraphim [82]

Narrow AI is the type of Artificial Intelligence (AI) can repeatedly perform tasks of limited scope.

  • Natural Language Understanding (NLU) is used to describe extracting information from unstructured text using algorithms.

<h3>What is Narrow AI ?</h3>

This is known to be a form or a type of AI that is said to be able to carry out a dedicated task as a result of its intelligence.

Note that Narrow AI is the type of Artificial Intelligence (AI) can repeatedly perform tasks of limited scope.

  • Natural Language Understanding (NLU) is used to describe extracting information from unstructured text using algorithms.

Learn more about Artificial Intelligence from

brainly.com/question/25523571

#SPJ1

7 0
2 years ago
What did the discovery of the Cumberland Gap mean for exploration? PLEASE HELP ILL GIVE YOU BRAINLEIST!
Flauer [41]

Answer: D. Many more people crossed the mountains and settled between the Appalachians and the Mississippi

Explanation:

It was quite difficult to cross the Appalachian mountains at the time of the American colonies which led to the colonies being confined to the east of it.

This was until the late 1700s when Daniel Boone and a team of explorers heard that a path existed through the Appalachians and had been used by natives for a long time. After discovering this ''gap'' which was then named after an English prince, many more people were able to use it to settle more to the west of the continent.

6 0
3 years ago
Consider an ideal gas undergoing a constant pressure process from state 1 to state
Radda [10]

Answer:

s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})

Explanation:

Hello,

In this case by combining the first and second law of thermodynamics for this ideal gas, we can obtain the following expression for the differential of the specific entropy <em>at constant pressure</em>:

ds=c_p\frac{dT}{T}-Rg\ \frac{dP}{P}

Whereas Rg is the specific ideal gas constant for the studied gas; thus, integrating:

\int\limits^{s_2}_{s_1} {} \, ds=c\int\limits^{T_2}_{T_1} {T^{d-1}dT} \,-Rg\ \int\limits^{P_2}_{P_1} {\frac{dP}{P}} \,

We obtain the expression to compute the specific entropy change:

s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})

Best regards.

6 0
4 years ago
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