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notsponge [240]
3 years ago
14

The thermometer reading 70◦Fis placed in an oven preheated to a constant temperature. Through the glass window in the oven door,

an observer records that the thermometer reads 110◦F after 12 minutes and 145◦F after 1 minute. How hot is the oven? Newton’s law of cooling yields the following differential equationdTdt=k(T−T0), whereT0 is the ambient temperature.
Physics
1 answer:
ValentinkaMS [17]3 years ago
5 0

Answer:

T0=390 degF

Explanation:

The thermometer reading 70◦Fis placed in an oven preheated to a constant temperature. Through the glass window in the oven door, an observer records that the thermometer reads 110◦F after 12 minutes and 145◦F after 1 minute. How hot is the oven? Newton’s law of cooling yields the following differential equationdTdt=k(T−T0), whereT0 is the ambient temperature.

If T is temperature, then by Newton’s law

dTdt=-k(T−T0)

Where

T0− is temperature of oven  

∫\frac{dT}{T-T0} =\int\limits^0.5_0 {-k} \, dt

ln(T-T0), from 110 to 70=-kt, from 1/2 to 0

ln(110-T0)-ln(70-T0)=-k(1/2-0)

2in\frac{110-T0}{70-T0} =-k\\k=-2ln\frac{T0-110}{T0-70} ...........................1

integrating the LHS of the equation from 110 to 145F

∫∫∫\frac{dT}{T-T0} =\int\limits^1_0.5 {-k} \, dt

ln(145-T0)-ln(110-T0)=-k(1-1/2)

2ln145-T0/(110-T0)=-k

k=-2lnT0-145/(T0-110).......................2

couplijng equatiuon 2 with 1

(T0-110)^2=(T0-70)(T0-145)

T0^2-220T0+12100=T0^2-215T0+10150

5T0=1950

T0=390 degF

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Answer:

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Large triangle Projector up to the screen

         tan θ = y / L

For the small triangle. Projector up to the person

         tan θ = y₀ / (L-d)

The angle is the same, so we equate the two equations

         y₀ / (L -d) = y / L

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The distance from the screen (d), we look for it with kinematics

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        d = v t

we replace

         y = y₀ L / (L - v t)

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         y = 114.4 (22 - 3t)⁻¹

This is the equation of the shadow height change as a function of time

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7 0
3 years ago
Describe two reasons why an alpha particle is less penetrating than a beta or gamma particle.
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8 0
3 years ago
Read 2 more answers
When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a
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Answer:

The right solution is:

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(b) 1.4375 eV

Explanation:

Given:

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= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = \frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}

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or,

  = \frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}

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(b)

As we know,

⇒ Vq=\frac{hc}{\lambda}-\Phi_0

By substituting the values, we get

⇒ 1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0

⇒                       \Phi_0=2.3\times 10^{-19} \ J

or,

⇒                            =\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}

⇒                            =1.4375 \ eV

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\\

hope helpful ~

8 0
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