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2 years ago

The layers in a high-mass star occur in order of:________

Physics

1 answer:

zavuch27 [327]2 years ago

6 0

A high-mass star has layers that are arranged in order of b. Temperature for fusion.

<h3>What is the temperature of fusion?</h3>

The temperature at which a substance transforms from a solid to a liquid is known as its melting point. The solid and liquid phases are in equilibrium at the melting point. Pressure affects a substance's melting point, which is typically reported at a standard pressure such 1 atmosphere or 100 kPa. There are two main categories of fusion reactions: those that maintain the ratio of protons to neutrons and those that convert protons to neutrons. The solid-liquid and liquid-to-gas transitions are connected with the typical heats of fusion and vaporization, respectively, for each substance.

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Find the magnitude of this<br> vector:<br> 174 m<br> N<br> 188.4 m<br> HELP FAST

Tomtit [17]

Answer:

195.168 m

Explanation:

To find the magnitude of the vector you can use the Pythagorean Theorem since you have the height and base and the vector is really just the hypotenuse

Pythagorean Theorem:

$a^2+b^2=c^2$

Plug values in

$88.4^2+174^2=c^2$

Simplify

$7814.56+30276=c^2$

Add the two values

$38090.56=c^2$

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$195.168\approx195.168$

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2 years ago

The rate at which the earth's surface loses heat and keeps earth temperature at which living things can survive

agasfer [191]

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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 $T=30^\circ F$ :

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