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Wittaler [7]
3 years ago
8

Consider the diatomic molecule oxygen O2 which is rotating in the xy plane about the z axis passing through its center, perpendi

cular to its length. The mass of each oxygen atom is 2.66 × 10−26 kg, and at room temperature, the average separation distance between the two oxygen atoms is 2.2 × 10−10 m (treat the atoms as point masses). If the angular speed of the molecule about the z axis is 5.49 × 1012 rad/s, what is its rotational kinetic energy? Answer in units of J.
Physics
1 answer:
Brilliant_brown [7]3 years ago
3 0

Answer: 9.7*10^-21 J

Explanation:

Given

Mass of oxygen atom, m = 2.66*10^-26 kg

Separation distance between the two atoms, r = 2.2*10^-10 m

Angular speed, w = 5.49*10^12 rad/s

First, we start by finding our moment of Inertia, I, using the relation

I = 1/2mr²

I = 1/2 * 2.66*10^-26 * (2.2*10^-10)²

I = 1/2 * 1.2874*10^-45

I = 6.437*10^-46 kgm²

The Rotational Kinetic Energy is given by

KE = 1/2Iw²

Now, using this value of I, we plug it in the energy equation.

KE = 1/2 * 6.437*10^-46 * (5.49*10^12)²

KE = 1/2 * 1.94*10^-20

KE = 9.7*10^-21 J

Thus, the rotational kinetic energy is 9.7*10^-21 J

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To solve this problem we will apply the definition of the ideal gas equation, where we will clear the density variable. In turn, the specific volume is the inverse of the density, so once the first term has been completed, we will simply proceed to divide it by 1. According to the definition of 1 atmosphere, this is equivalent in the English system to

1atm = 2116lb/ft^2

The ideal gas equation said us that,

PV = nRT

Here,

P = pressure

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Then

\frac{n}{V} = \frac{P}{RT}

The amount of substance per volume is the density, then

\rho = \frac{P}{RT}

Replacing with our values,

\rho = \frac{5.6*2116}{1716*850}

\rho = 0.00812slug/ft^3

Finally the specific volume would be

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3 years ago
A shell is fired at an angle of 35° above the horizontal at a velocity of 40 m/s. (a) What is it's speed at the highest point of
Fudgin [204]

Answer:

Part a)

v_f = v_x = 32.77 m/s

Part b)

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Explanation:

Part a)

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so here we have

v_x = 40 cos35 = 32.77 m/s

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v_f = 32.77 m/s

Part b)

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so we have

\Delta y = 0

0 = v_y t - \frac{1}{2}gt^2

T = \frac{2v_y}{g}

T = \frac{2(22.94)}{9.81} = 4.68 s

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The speed of the water is the greatest at point B

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lilavasa [31]

Answer:

.409 N

Explanation:

For this to balance, the moments around the fulcrum must sum to zero.

On the left you have   .21   ( is that down? I will assume it is)

      Counterclockwise moments :

        .21 * 40     +  1.0 * 20    

     Clockwise moments :

        .5 * 20     +     F * 45

these moments must equal each other

.21*40 + 1 *20   =  .5 * 20 + F * 45

   F = .409 N

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Is it possible to put a distance versus time graph to be universal go mine
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