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Artemon [7]
2 years ago
15

At room temperature, an oxygen molecule, with mass of 5.31x10-26kg , typically has a kinetic energy of about 6.21x10-21J.How fas

t is it moving?
Physics
1 answer:
alex41 [277]2 years ago
3 0
<h2>Oxygen is moving at 483.63 m/s</h2>

Explanation:

Kinetic energy = 0.5 x Mass x Velocity²

Kinetic energy = 6.21 x 10⁻²¹ J

Mass = 5.31 x 10⁻²⁶ kg

Substituting

        Kinetic energy = 0.5 x Mass x Velocity²

        6.21 x 10⁻²¹ = 0.5 x 5.31 x 10⁻²⁶ x Velocity²

        Velocity² = 2.34 x 10⁵

        Velocity = 483.63 m/s

Oxygen is moving at 483.63 m/s  

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The pfsense firewall, like other firewalls on the market, relies on __________ to expose an ip address from the private network
sashaice [31]
T<span>he pfsense firewall, like other firewalls on the market, relies on the subnet mask to expose an ip address from the private network and bind it to an address on the public network. </span>
6 0
2 years ago
You are standing at the midpoint between two speakers, a distance D away from each. The speakers are playing the exact same soun
Rzqust [24]

Answer:

Explanation:

wave length of sound waves = velocity / frequency

= 340 / 170

λ = 2 m.

When the position of man is exactly at the meddle point between the speakers , sound waves from the speakers reaching man are in same phases ( path difference is zero. ) so intensity of sound is maximum .

Now , the man starts moving towards one of the speakers , his distance from one speaker becomes closer than the other creating path difference for the sound waves reaching his ears.

If he walks a distance of .5 m towards one speaker , path difference created

= .5 x 2 = 1 m

So , path difference = λ /2 ,

there will be destructive interference so minimum sound will be heard there.

When he walks a distance of 1 m , path difference created = 2m

path difference =  λ

so there is constructive interference and maximum  sound will be heard there.

Again we he walks a distance of 1.5 m , path difference created = 3 m

path difference = 3 λ /2

So there will be destructive interference so minimum sound will be heard there.

In this way we see that man starts  from a point of maximum sound intensity , reaches a point of minimum sound intensity , then reaches a point of maximum sound intensity . At last he reaches a position of minimum sound intensity.

3 0
2 years ago
In the sport of curling, large smooth stones are slid across an ice court to land on a target. Sometimes the stones need to move
lara31 [8.8K]

Answer:

To increase kinetic friction, the amount of fine water droplets sprayed before the game is limited.

To reduce kinetic friction. increase the amount of fine water droplets during pregame preparation and sweeping in front of the curling stones.

Explanation:

In curling sports, since the ice sheets are flat, the friction on the stone would be too high and the large smooth stone would not travel half as far. Thus controlling the amount of fine water droplets sprayed before the game is limited pregame is necessary to increase friction.

On the other hand, reducing ice kinetic friction involves two ways. The first way is adding bumps to the ice which is known as pebbling. Fine water droplets are sprayed onto the flat ice surface. These droplets freeze into small "pebbles", which the curling stones "ride" on as they slide down the ice. This increases contact pressure which lowers the friction of the stone with the ice. As a result, the stones travel farther, and curl less.  

The second way to reduce the kinetic friction is sweeping in front of the large smooth stone. The sweeping action quickly heats and melts the pebbles on the ice leaving a film of water. This film reduces the friction between the stone and ice.

8 0
2 years ago
In a semiclassical model of the hydrogen atom, the electron orbits the proton at a distance of 0.053 nm. Part A What is the elec
Bezzdna [24]

Answer with Explanation:

We are given that

r=0.053 nm=0.053\times 10^{-9} m

1 nm=10^{-9} m

Charge on proton,q=1.6\times 10^{-19} C

a.We have to find the electric  potential of the proton at the position of the electron.

We know that the electric potential

V=\frac{kq}{r}

Where k=9\times 10^9

V=\frac{9\times 10^9\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

V=27.17 V

B.Potential energy of electron,U=\frac{kq_e q_p}{r}

Where

q_e=-1.6\times 10^{-19} c=Charge on electron

q_p=q=1.6\times 10^{-19} C=Charge on proton

Using the formula

U=\frac{9\times 10^9\times (-1.6\times 10^{-19}\times 1.6\times 10^{-19}}{0.053\times 10^{-9}}

U=-4.35\times 10^{-18} J

8 0
2 years ago
A circular surface with a radius of 0.057 m is exposed to a uniform external electric field of magnitude 1.44 × 104 N/C. The mag
klio [65]

Answer:

57.94°

Explanation:

we know that the expression of flux

\Phi =E\times S\times COS\Theta

where Ф= flux

           E= electric field

           S= surface area

        θ = angle between the direction of electric field and normal to the surface.

we have Given Ф= 78 \frac{Nm^{2}}{sec}

                          E=1.44\times 10^{4}\frac{Nm}{C}

                          S=\pi \times 0.057^{2}

                         COS\Theta =\frac{\Phi }{S\times E}

 =   \frac{78}{1.44\times 10^{4}\times \pi \times 0.057^{2}}

 =0.5306

 θ=57.94°

4 0
2 years ago
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