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2 years ago

The first model of the atom was developed through

Physics

2 answers:

Ilya [14]2 years ago

7 0

It was developed through Democritus who was a greek philosopher.

Hope this helps

padilas [110]2 years ago

3 0

The correct answer is C. thinking about the smallest particles of matter without experimenting.

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How many values must a 1-D<br> motion problem give you in<br> order to solve it?

MrRa [10]

Answer:

I think it is B: 3

Explanation:

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2 years ago

Why cooking pans don't have have metal handles?

Harrizon [31]

metals are conductors, meaning that heat travels through it better than insulators such as plastic (commonly used for pan handles)

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2 years ago

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A child sits 1.6 m from the center of a merry go round that makes one complete revolution in 4.7 s. What is his angular accelera

olga2289 [7]

Answer: angular acceleration = $2.86\ rad/sec^{2}$

Given:

Distance from center of axis = 1.6 m

Time taken to complete one revolution = 4.7 sec

Therefore, we can evaluate the angular acceleration using the following formula:

$\alpha = r\times \omega^{2}$

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$\alpha = 1.6\times (\frac{2\times3.14}{4.7})^{2}$

$\alpha$ = $2.86\ rad/sec^{2}$

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3 years ago

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2 years ago

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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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3 years ago