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3 years ago

Which factors affect heat transfer between a warm and a cool substance?

Physics

1 answer:

andrey2020 [161]3 years ago

5 0

Answer: B

Explanation:

It's not the time it took to heat the substance, so that rules out A and C.

This means that we only have to choose between

B. the area of contact

D. the area of the substances

(since everything else in each of those answers are the same)

Area of contact matters more (e.g. an object with greater surface area is exposed to the air more, will lose/gain heat quicker than an object with less surface area).

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Lagrangian mechanics. Determine the equations of motion for a particle of mass m constrained to move on the surface of a cone in

maria [59]

Answer:

Explanation:

Hi!

In order to obtain the Lagrangian of the system we must first write the Kinetic and Potential Energies. Lets orient our axes such that the axis of the cone coincide with the z axis. In cilindrical coordinates we have

$v^{2} = \frac{dr}{dt}^{2} +r^{2} \frac{d\theta }{dt} ^{2} +\frac{dz}{dt} ^{2}$ - (1)

But, since the particle is constrained to move on the surface of the cilinder, we have the following relation between r and z:

$\frac{r}{z}=tan(45)$

or:

$z = r cot(45)$ - (2)

and:

$\frac{dz}{dt} = \frac{dr}{dt} cot(45)$

replacing (2) in (1) we obtain:

$v^{2} = \frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2}$ - (3)

Now the kinetic energy is given as:

$T = \frac{1}{2}m(\frac{dr}{dt}^{2} (1+cot(45))+r^{2}\frac{d\theta }{dt} ^{2})$ - (4)

And the potential energy is given by:

$V = -mgz = -mgr cot(45)$

So the Langrangian is given by:

$L = T - V= \frac{1}{2}m(\frac{dr}{dt}^{2}(1+cot(45)+r^{2})\frac{d\theta }{dt} ^{2}) + mgr cot(45)$

And the equations of motion are:

For θ

$\frac{d}{dt} (mr\frac{d\theta}{dt}) = 0-->mr{d\theta}{dt}=c$

For r

$\frac{d}{dt}(m\frac{dr}{dt}(1+cot(45) )= mgcot(45)+mr\frac{d\theta}{dt} ^{2}\\m\frac{d^{2} r}{dt^{2} }(1+cot(45)= mgcot(45)+mr\frac{d\theta}{dt} ^{2}$

Obtained from the Euler-Langrange equations

Here the conserved quantity is given by the first equation of motion, namely:

$mr\frac{d\theta}{dt}=c$

Which is the magnitude of the angular momentum

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3 years ago

When the cart moves on an incline at constant speed, it is in equilibrium; i.e., the net force on it is zero. Does it require mo

Amanda [17]

Answer:

It requires more tension to pull up the track

Explanation:

Net force must be zero to maintain constant velocity.

Weight force will always be pointed down the slope. Call it W

Friction force (Call it Ff) will be down slope when movement is up slope.

Friction force will be up slope when movement is down slope.

W and Ff are always positive numbers

Call the pulling force T

If Up slope is considered the positive direction

Moving up slope

Tu - Ff - W = 0

Tu = Ff + W

Moving down slope

Td + W - Ff = 0

Td = Ff - W

Ff + W > Ff - W therefore Tu > Td

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2 years ago

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Furkat [3]

Answer: Chlorophyll Breaking down.

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