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2 years ago

Fomula for atmospheric presure

Physics

1 answer:

katovenus [111]2 years ago

3 0

Answer:

P_h= P_0 e^{\frac {-mgh}{kT}}

Explanation:

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What do you want to learn about simple machines? (could someone come up with a question about simple machines for me)

bezimeni [28]

Answer:

what are simple machines lol

Explanation:

6 0

2 years ago

An amusement park ride consists of a rotating circular platform 11.1 m in diameter from which 10 kg seats are suspended at the e

frozen [14]

To solve this problem we will use the relationship given between the centripetal Force and the Force caused by the weight, with respect to the horizontal and vertical components of the total tension given.

The tension in the vertical plane will be equivalent to the centripetal force therefore

$Tsin\theta= \frac{mv^2}{r}$

Here,

m = mass

v = Velocity

r = Radius

The tension in the horizontal plane will be subject to the action of the weight, therefore

$Tcos\theta = mg$

Matching both expressions with respect to the tension we will have to

$T = \frac{\frac{mv^2}{r}}{sin\theta}$

$T = \frac{mg}{cos\theta}$

Then we have that,

$\frac{\frac{mv^2}{r}}{sin\theta} = \frac{mg}{cos\theta}$

$\frac{mv^2}{r} = mg tan\theta$

Rearranging to find the velocity we have that

$v = \sqrt{grTan\theta}$

The value of the angle is 14.5°, the acceleration (g) is 9.8m/s^2 and the radius is

$r = \frac{\text{diameter of rotational circular platform}}{2} + \text{length of chains}$

$r = \frac{11.1}{2}+2.41$

$r = 7.96m$

Replacing we have that

$v = \sqrt{(9.8)(7.96)tan(14.5\°)}$

$v = 4.492m/s$

Therefore the speed of each seat is 4.492m/s

6 0

3 years ago

Some element can be either solid or liquid. At the melting point, the liquid has 8 × 10-22 J more enthalpy per atom than the sol

OlgaM077 [116]

Answer:

481.76 J/mol

133.33 K

Explanation:

$N_A$ = Avogadro's number = $6.022\times 10^{23}$

Change in enthalpy is given by

$\Delta H=8\times 10^{-22}\times 6.022\times 10^{23}\\\Rightarrow \Delta H=481.76\ J/mol$

Entropy is given by

$\Delta S=6\times 10^{-24}\times 6.022\times 10^{23}\\\Rightarrow \Delta S=3.6132\ J/mol K$

Latent heat of fusion is given by

$L_f=\Delta H\\\Rightarrow L_f=481.76\ J/mol$

The latent heat of fusion is 481.76 J/mol

Melting point is given by

$T_m=\dfrac{L_f}{\Delta S}\\\Rightarrow T_m=\dfrac{8\times 10^{-22}\times 6.022\times 10^{23}}{6\times 10^{-24}\times 6.022\times 10^{23}}\\\Rightarrow T_f=133.33\ K$