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vichka [17]
3 years ago
9

How is thermal energy transferred during convection? by physical contact by electromagnetic waves by movement in fluids by mecha

nical waves
Physics
2 answers:
Alla [95]3 years ago
7 0

by movement in fluids

iris [78.8K]3 years ago
4 0

Answer: by movement in fluids

Explanation:

There are three modes of transfer of thermal energy.

Conduction, convection and radiation.

Transfer of thermal energy via <u>conduction</u> takes place from hotter object and to colder object when they are contact.

Transfer of thermal energy via convection takes place through fluids. The bulk motion of fluids transfers energy.

<u>Radiation </u>transfers thermal energy via space.

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A tortoise can run with a speed of 0.14 m/s, and a hare can run 20 times as fast. In a race, they both start at the same time, b
Reptile [31]

speed of tortoise is given as v1 = 0.14 m/s

speed of hare is given as v2 = 20*0.14 = 2.8 m/s

now let say the total length of the path is "d"

so the total time taken by the tortoise to cover this

t = \frac{d}{0.14}

now given that hare took rest for 1 min

so total time of run for hare is (t - 60)s

so the distance that hare covered is given by

d - 0.30 = 2.8 * (t - 60)

now by above two equations

d - 0.30 = 2.8 * \frac{d}{0.14} - 168

168 - 0.30 = (20 - 1)d

d = 8.82 m

and the time t is given by

t = \frac{8.82}{0.14}

t = 63 s

so part a)

t = 63 s

part b)

d = 8.82 m

4 0
3 years ago
Consider a semi-infinite (hollow) cylinder of radius R with uniform surface charge density. Find the electric field at a point o
VikaD [51]

Answer:

For the point inside the cylinder: E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}

For the point outside the cylinder: E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}

where x0 is the position of the point on the x-axis and σ is the surface charge density.

Explanation:

Let us assume that the finite end of the cylinder is positioned at the origin. And the rest of the cylinder lies on the (-x) axis, which is the vertical axis in this question. In the first case (inside the cylinder) we will calculate the electric field at an arbitrary point -x0. In the second case (outside), the point will be +x0.

<u>x = -x0:</u>

The cylinder is consist of the sum of the rings with the same radius.

First we will calculate the electric field at point -x0 created by the ring at an arbitrary point x.

We will also separate the ring into infinitesimal portions of length 'ds' where ds = Rdθ.

The charge of the portion 'ds' is 'dq' where dq = σds = σRdθ. σ is the surface charge density.

Now, the electric field created by the small portion is 'dE'.

dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2}

The electric field is a vector, and it needs to be separated into its components in order us to integrate it. But, the sum of horizontal components is zero due to symmetry. Every dE has an equal but opposite counterpart which cancels it out. So, we only need to take the component with the sine term.

dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2} \frac{x}{\sqrt{x^2+R^2}} = dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rxd\theta}{(R^2+x^2)^{3/2}}

We have to integrate it over the ring, which is an angular integration.

E_{ring} = \int{dE} = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta  = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}2\pi = \frac{1}{2\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}

This is the electric field created by a ring a distance x away from the point -x0. Now we can integrate this electric field over the semi-infinite cylinder to find the total E-field:

E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{-2x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}

The reason we integrate over -2x0 to -inf is that the rings above -x0 and below to-2x0 cancel out each other. Electric field is created by the rings below -2x0 to -inf.

<u>x = +x0: </u>

We will only change the boundaries of the last integration.

E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}

6 0
3 years ago
You are a support technician working in a data closet in a remote office. You suspect that a connectivity problem is related to
Nostrana [21]

Answer:

crimping tool

Explanation:

This is a tool employed in affixing a connector to the end of a network cable.

5 0
3 years ago
A supercapacitor is an electrical energy storage device. A supercapacitor, initially charged to 2.1 thousand millivolts, supplie
svet-max [94.6K]

Answer:

the time taken t is 9.25 minutes

Explanation:

Given the data in the question;

The initial charge on the supercapacitor = 2.1 × 10³ mV = 2.1 V

now, every minute, the charge lost is 9.9 %  

so we need to find the time for which the charge drops below 800 mV or 0.8 V

to get the time, we can use the formula for compound interest in basic mathematics;

A = P × ( (1 - r/100 )ⁿ

where A IS 0.8, P is 2.1, r is 9.9

so we substitute

0.8 = 2.1 × ( 1 - 0.099 )ⁿ

0.8/2.1 = 0.901ⁿ

0.901ⁿ = 0.381

n = 9.25 minutes

Therefore, the time taken t is 9.25 minutes

6 0
3 years ago
If a person looking at a poster sees green instead of yellow and doesn't see red at all, this person most likely has color blind
katrin [286]
Red - sensitive so answer is c
7 0
3 years ago
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