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rodikova [14]
3 years ago
14

Which of the following is likely to have the lowest viscosity? hot oil room temperature oil room temperature water below room te

mperature oil
Hot oil
Room temperature oil
Room temperature water
Below room temperature oil
Physics
2 answers:
RoseWind [281]3 years ago
6 0

Answer: Option (a) is the correct answer.

Explanation:

Viscosity is defined as the inability of a liquid to flow easily. Basically, a liquid which is very thick will have the highest viscosity.

Molecules of some liquids are more closely packed due to which they face resistance in movement from one place to another.

On the other hand, when we heat a liquid then molecules gain kinetic energy. As a result, they tend to move away from each other. This movement of molecules make the flow of liquid easy and it will have the lowest viscosity.

Thus, we can conclude that out of the following hot oil is likely to have the lowest viscosity.

Zanzabum3 years ago
5 0
The answer should be hot oil

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A sand mover at a quarry lifts 2,000 kg of sand per minute a vertical distance of 12 m. The sand is initially at rest and is dis
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Answer:

<h2>E. 3.95kW</h2>

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3 years ago
A force acting over a large area will exert less pressure per square inch than the same force acting over a smaller area.
babunello [35]

Answer:

True

Explanation:

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p=\frac{F}{A}

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p\propto \frac{1}{A}

this means that, assuming that the forces in the two situations (which have same magnitude) are both applied perpendicular to the surface, the force exerted over the smaller area will exert a greater pressure. Hence, the statement"

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Read 2 more answers
A flywheel with a diameter of 1.42 m is rotating at an angular speed of 207 rev/min. (a) What is the angular speed of the flywhe
Archy [21]

Answer:

a. 21.68 rad/s b. 30.78 m/s c. 897 rev/min² d. 1085 revolutions

Explanation:

a. Its angular speed in radians per second  ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s

b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m

So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s

c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min

α = (ω₁ - ω)/t

  = (1410 - 207)/(80.5/60)

  = 60(1410 - 207)/80.5

  = 60(1203)80.5

  = 896.65 rev/min² ≅ 897 rev/min²

d. Using θ = ωt + 1/2αt²

where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and  t = 80.5/60 min = 1.342 min

θ = ωt + 1/2αt²

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  = 277.725 + 807.417

  = 1085.14 revolutions ≅ 1085 revolutions

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Ede4ka [16]

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