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Stella [2.4K]
3 years ago
13

Jimmy is cooking pasta in a pot of water on the stove. He uses a stopwatch to see how long the pasta takes to cook depending on

the temperature of the water in the pot. What is the independent variable and why?
Physics
1 answer:
Alex787 [66]3 years ago
4 0

Answer:

stove

Explanation:

The stove is the independent variable in this question as it does not depend on anything like other variables such as pot, water, temperature and of cause pasta.

The pot depends on the stove, the water depends on the pot and stove, the pasta depends on the water and its temperature.

Cheers.  

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Answer:

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3 years ago
A thermometer is removed from a room where the temperature is 70° F and is taken outside, where the air temperature is 10° F. Af
vekshin1

Answer:

T=51.64^\circ F

t=180.10s

Explanation:

The Newton's law in this case is:

T(t)=T_m+Ce^{kt}

Here, T_m is the air temperture, C and k are constants.

We have

70^\circ F in t=0

So:

T(0)=70^\circ F\\T(0)=10^\circ F+Ce^{k(0)}\\70^\circ F=10^\circ F+C\\C=70^\circ F-10^\circ F=60^\circ F

And we have 60^\circ F in t=30 s, So:

T(30)=60^\circ F\\T(30)=10^\circ F+(60^\circ F)e^{k(30)}\\60^\circ F=10^\circ F+(60^\circ F)e^{k(30)}\\50^\circ F=(60^\circ F)e^{k(30)}\\e^{k(30)}=\frac{50^\circ F}{60^\circ F}\\(30)k=ln(\frac{50}{60})\\k=\frac{ln(\frac{50}{60})}{30}=-0.0061

Now, we have:

T=10^\circ F+(60^\circ F)e^{-0.0061t}(1)

Applying (1) for t=1 min=60s:

T=10^\circ F+(60^\circ F)e^{-0.0061*60}\\T=10^\circ F+(60^\circ F)0.694\\T=10^\circ F+41.64^\circ F\\T=51.64^\circ F

Applying (1) for T=30^\circ F:

30^\circ F=10^\circ F+(60^\circ F)e^{-0.0061t}\\30^\circ F-10^\circ F=(60^\circ F)e^{-0.0061t}\\-0.0061t=ln(\frac{20}{60})\\t=\frac{ln(\frac{20}{60})}{-0.0061}=180.10s

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Answer:

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Answer:

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Do you have multiple choice i can see?

Explanation:

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What is the wavelength that corresponds to a frequency of 6.00x1014 hz?
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