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kirza4 [7]
3 years ago
10

Oil flows upward in the wick of a lantern because of the liquid property called

Physics
1 answer:
aalyn [17]3 years ago
4 0
Capillarity  The state of being capillary; the attraction and repulsion of molecules on the surface of a liquid in contact with a solid, as the walls of it's container ,which results in the rise or fall of a liquid when in narrow tubes or fibers, or in the wetting of a solid by a liquid.
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What do scientist do to receive suggestions an criticism of their research from other scientists?
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They PUBLISH !

They provide complete written, detailed, technical descriptions of their thoughts, hypotheses, experiments, data, and conclusions, in publications that are read by other scientists around the world.

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4 years ago
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Define torque qualitatively and quantitatively.
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When you use a wrench to tighten or loosen a nut on a bolt, you are applying torque. It is measured in units of force times distance.  A force of F newtons pulling on a handle of L meters in length would supply a torque of F L newton-meters. More technically, torque is the vector cross product of force times perpendicular distance from the object, F x r = F r sin @
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3 years ago
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7. A toy car of mass 1.2 kg is driving vertical circles inside a hollow cylinder of radius 2.0m. It is moving at a constant spee
wlad13 [49]

Answer:

a)

N_{top}=9.8N\\N_{bottom}=33.4N

b) v_{min}=4.4m/s

Explanation:

The net force on the car must produce the centripetal acceleration necessary to make this circle, which is a_{cp}=\frac{v^2}{R}. At the top of the circle, the normal force and the weight point downwards (like the centripetal force should), while at the bottom the normal force points upwards (like the centripetal force should) and the weight downwards, so we have (taking the upwards direction as positive):

-m\frac{v^2}{R}=-N_{top}-mg\\m\frac{v^2}{R}=N_{bottom}-mg

Which means:

N_{top}=m\frac{v^2}{R}-mg=(1.2kg)\frac{(6m/s)^2}{2m}-(1.2kg)(9.8m/s^2)=9.8N\\N_{bottom}=m\frac{v^2}{R}+mg=(1.2kg)\frac{(6m/s)^2}{2m}+(1.2kg)(9.8m/s^2)=33.4N

The limit for falling off would be N_{top}=0, so the minimum speed would be:

0=m\frac{v_{min}^2}{R}-mg\\v_{min}=\sqrt{Rg}=\sqrt{(2m)(9.8m/s^2)}=4.4m/s

3 0
3 years ago
Consider a vortex filament of strength Γ in the shape of a closed circular loop of radius R. Consider also a straight line throu
Sedbober [7]

Answer:

\vec{V} = \frac{\Gamma}{2R}\vec{A}

Explanation:

We define our values according to the text,

R= Radius

\vec{V} =Velocity

\Gamma =Strenght of the vortex filament

From this and in a vectorial way we express an elemental lenght of this filmaent as \vec{dl}. So,

\vec{dl}x\vec{r} = R*dl*\vec{A}

Where \vec{A} imply a vector acting perpendicular to both vectors.

Applying Biot-Savart law, we have,

\vec{V} =\frac{\Gamma}{4\pi}\int\frac{\vec{dl}x\vec{r}}{r^3}

Substituting the preoviusly equation obtained,

\vec{V} = \frac{\Gamma}{4\pi}\int\frac{R*dl*\vec{A}}{R^3}

\vec{V} = \frac{\Gamma}{4\pi R^2}\int^{2\pi R}_0 dl*\vec{A}

\vec{V} = \frac{\Gamma(2\pi R \vec{A})}{4\pi R^2}

So we can express the velocity induced is,

\vec{V} = \frac{\Gamma}{2R}\vec{A}

6 0
3 years ago
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