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3 years ago

7

The furnace keeps houseAat 25◦C, while thefurnace in houseBkeeps it at 20◦C. Which house requires heat to be supplied by its fur

nace at a faster rate?
1.House B requires heat at a slightly faster rate than A.
2.House A requires heat at about twice the rate of B.
3.House A requires heat at about four times the rate of B.
4.Houses A and B require heat at the same rate.
5.House A requires heat at a slightly faster rate than B.
6.House B requires heat at about twice the rate of A.
7.House B requires heat at about four times the rate of A.
8.None of the above answers is correct.

1 answer:

EleoNora [17]3 years ago

5 0

Answer:

House A requires heat at a slightest faster rate than B

Explanation:

House A requires heat at a slightest faster rate than B due to the slight high temperature the furnace A is.

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According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave lengt

scZoUnD [109]

Answer:1.55 times

Explanation:

Given

First wavelength $(\lambda _1)=450 nm$

Second wavelength $(\lambda _2)=700 nm$

According wien's diplacement law

$\lambda T=constant$

where $\lambda =wavelength$

T=Temperature

Let $T_1 and T_2$ be the temperatures corresponding to $\lambda _1 & \lambda _2$ respectively.

$\lambda _1\times T_1=\lambda _2\times T_2$

$\frac{T_1}{T_2}=\frac{\lambda _2}{\lambda _1}$

$\frac{T_1}{T_2}=\frac{700}{450}=1.55$

Thus object with $\lambda 450 nm$ is 1.55 times hotter than object with wavelength $\lambda =700 nm$

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3 years ago

Carbon is allowed to diffuse through a steel plate 15 mm thick. The concentrations of carbon at the two faces are 0.65 and 0.30

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Answer:

T=575.16K

Explanation:

To solve the problem we proceed to use the 1 law of diffusion of flow,

Here,

$J=-D\frac{\Delta C}{\Delta x}$

$\Delta C$ is the rate in concentration

$\Delta x$ is the rate in thickness

D is the diffusion coefficient, where,

$D= D_0 exp(\frac{Q_d}{RT})$

Replacing D in the first law,

$J=-(D_0 exp(\frac{-Q_D}{RT}))\frac{\Delta }{\Delta x}$

clearing T,

$T=\frac{Q_d}{R*ln(\frac{J*\Delta x}{D_0*\Delta C})}$

Replacing our values

$T=-\frac{80000}{8.31*ln(\frac{(6.2*10^{-7})(-15*10^{-3})}{(1.43*10^{-9})(0.65-0.30)})}$

$T=-\frac{80000}{-138.09}$

$T=575.16K$

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A pendulum is not a wave.

-- A pendulum doesn't have a 'wavelength'.

-- There's no way to define how many of its "waves" pass a point
every second.

-- Whatever you say is the speed of the pendulum, that speed
can only be true at one or two points in the pendulum's swing,
and it's different everywhere else in the swing.

-- The frequency of a pendulum depends only on the length
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If you take the given information and try to apply wave motion to it:

   Wave speed = (wavelength) x (frequency)

   Frequency = (speed) / (wavelength) ,

you would end up with

   Frequency = (30 meter/sec) / (0.35 meter) = 85.7 Hz

Have you ever seen anything that could be described as
a pendulum, swinging or even wiggling back and forth
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This math is not applicable to the pendulum.

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I hope this helps ~~Charlotte~~

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3 years ago

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