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3 years ago

5

The transfer of heat by the movement of a fluid is called: Think about it: Imagine a pocket of air over the land (“land air”), a

nd another pocket of air over the ocean (“ocean air”).
Which air pocket would you expect to heat up more during the day?

Why?

1 answer:

timama [110]3 years ago

7 0

Answer:

The transfer of heat by the movement of fluid is called Convection Heat Transfer

Explanation:

Heat transfer by convection is the transfer of heat by fluid transport from one place to another, such that convection takes place when the heat that comes in contact of fluid containing body is moved to other parts of the container by the transporting fluid

Heat is transferred within a fluid medium mainly by convection (movement of heat by the transfer of fluid particles in the medium)

Convection heat transfer is a combination of conduction and advection heat transfer

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My mom and I walked 3,600 m in 90 minutes. What was our speed in m/min?

PtichkaEL [24]

Fill in the fraction: 3,600/90 = 40; turn it into a unit fraction.

40 mi/min

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3 years ago

A proposed space station consists of a circular tube that will rotate about its center (like a tubular bicycle tire), Fig. 5–39.

Trava [24]

Answer:

1742.24106 revolutions per day

Explanation:

v = Velocity

d = Diameter = 1.1 km

r = Radius = $\dfrac{d}{2}=\dfrac{1.1}{2}=0.55\ km$

g = Acceleration due to gravity = 9.81 m/s²

g = 0.9 g

The centrifugal force will balance the gravitational force

$F_c=mg\\\Rightarrow \dfrac{mv^2}{r}=m0.9g\\\Rightarrow v=\sqrt{\dfrac{0.9gmr}{m}}\\\Rightarrow v=\sqrt{0.9gr}\\\Rightarrow v=\sqrt{0.9\times 9.81\times 0.55\times 10^3}\\\Rightarrow v=69.68464\ m/s$

$\dfrac{1}{T}=\dfrac{v}{2\pi r}\\\Rightarrow \dfrac{1}{T}=\dfrac{69.68464}{2\pi 0.55\times 10^3}\times 24\times 60\times 60\\\Rightarrow \dfrac{1}{T}=1742.24106\ rev/day$

The rotation speed is 1742.24106 revolutions per day

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3 years ago

Read 2 more answers

Two objects gravitationally attract with a force of 10N. If the distance between the two objects center is doubled, then the new

lana [24]

Answer:

20N

Explanation:

10×2

5 0

3 years ago

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

<h3>a.</h3>

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

<h3>b.</h3>

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

<h2>b</h2>

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

3 0

3 years ago

Two balls of unequal mass are hung from two springs that are not identical. The springs stretch the same distance as the two sys

IRISSAK [1]

Answer:

b) Springs oscillate with the same frequency,

Explanation:

expression for frequency of vibration of mass hanging from a spring is given as follows

f = $\frac{1}{2\pi} \times \sqrt{\frac{k}{m} }$

k is force constant of spring and m is mass vibrating .

In the present case, if mass stretches the spring by x and remains balanced

mg = k x

$\frac{k}{m} =\frac{g}{x}$

g and x are same for both cases

$\frac{k}{m}$ will also be same for both cases .

Hence frequency of vibration will also be same for both the balls .

4 0

3 years ago