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Kryger [21]
3 years ago
9

ANSWER QUICK 30 POINTS

Physics
1 answer:
Lelu [443]3 years ago
4 0
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For a molecule of O2 at room temperature (300 K), calculate the average angular velocity for rotations about the x or y axes. Th
Ksivusya [100]

Answer: 6.65 rad/s

Explanation:

Firstly, we need to know that according the kinetic theory of gases, the average kinetic energy KE of a molecule with i degrees of freedom is:

KE=\frac{i}{2}kT (1)

Where:

i=5 because oxigen is a diatomic molecule and has 5 degrees of freedom

k=1.38064852 (10)^{-23} \frac{m^{2}kg}{s^{2}K} is the Boltzman constant

T=300 K is the temperature

Then:

KE=\frac{5}{2}(1.38064852 (10)^{-23} \frac{m^{2}kg}{s^{2}K})(300 K) (2)

KE=1.035 (10)^{-20} \frac{m^{2}kg}{s^{2}} (3) With this value of average kinetic energy we can find the average angular velocity, with the following equation:

KE=\frac{1}{2}I \omega^{2} (4)

Where:

I is the moment of inertia

\omega is the angular velocity

Now, I=m R^{2} (5)

Being m=31.98 g/mol=0.03198 kg/mol the molar mass of Oxigen molecule and R=0.121(10)^{-9}m the distance between atoms

I=(0.03198 kg/mol)(0.121(10)^{-9}m)^{2} (6)

I=4.682(10)^{-22} \frac{kg}{mol} m^{2} (7)

Substituting (7) and (3) in (4):

1.035 (10)^{-20} \frac{m^{2}kg}{s^{2}}=\frac{1}{2} (4.682(10)^{-22} \frac{kg}{mol} m^{2}) \omega^{2} (8)

Finding \omega:

\omega=6.65 rad/s (9) This is the average angular velocity for a molecule of O2

7 0
3 years ago
Which list includes the phase changes that require a loss of energy (heat)?
Lemur [1.5K]

Answer:

b  

Explanation:

because if you freeze somthing you do not gain heat you loss heat.

3 0
2 years ago
If the atoms that share electrons have an unequal attraction for the electrons is called
Karolina [17]

If the atoms that share electrons have an unequal attraction for electrons, the bond is called a Polar covalent bond.

<h3><u>Explanation:</u></h3>

A covalent chemical bond is formed in case of two different non-metals when one or more electron pairs are shared between bonding atoms. A difference in electronegativity of subsequent atoms of a covalent bond leads to formation of a small net charge around nucleus of each atom, pulling the shared electrons to one side of the bond, to the nucleus which has higher electronegativity.

HCl is an example of polar covalent bond and the HCl bond has Chlorine more electronegative. The bonding electrons are more close to Cl than H and hence Cl is partially negatively charged than H which has partial positive charge (HCl bond : H^{+} - Cl^{-}). When electrons shared in a covalent bond have equal attraction, the bond is a Non-Polar covalent bond.

6 0
3 years ago
An athlete swings a ball, connected to the end of a chain, in a horizontal circle. The athlete is able to rotate the ball at the
garik1379 [7]

(a) 6.04 rev/s

The speed of the ball is given by:

v=\omega r

where

\omega is the angular speed

r is the distance of the ball from the centre of the circle

In situation 1), we have

\omega=8.13 rev/s \cdot 2\pi = 51.0 rad/s

r = 0.600 m

So the speed of the ball is

v=(51.0 rad/s)(0.600 m)=30.6 m/s

In situation 2), we have

\omega=6.04 rev/s \cdot 2\pi = 37.9 rad/s

r = 0.900 m

So the speed of the ball is

v=(37.9 rad/s)(0.900 m)=34.1 m/s

So, the ball has greater speed when rotating at 6.04 rev/s.

(b) 1561 m/s^2

The centripetal acceleration of the ball is given by

a=\frac{v^2}{r}

where

v is the speed

r is the distance of the ball from the centre of the trajectory

For situation 1),

v = 30.6 m/s

r = 0.600 m

So the centripetal acceleration is

a=\frac{(30.6 m/s)^2}{0.600 m}=1561 m/s^2

(c) 1292 m/s^2

For situation 2 we have

v = 34.1 m/s

r = 0.900 m

So the centripetal acceleration is

a=\frac{v^2}{r}=\frac{(34.1 m/s)^2}{0.900 m}=1292 m/s^2

5 0
3 years ago
A jogger runs 5.0 km on a straight trail at an angle of 60° south of west. What is the southern component of the run rounded to
geniusboy [140]

Answer:

4.3 km

Explanation:

5 0
2 years ago
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