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uranmaximum [27]
2 years ago
12

Why are ionized gases typically only found in very high-temperature environments?

Physics
1 answer:
LenaWriter [7]2 years ago
7 0

Explanation:

<u>Ionized gases are found in very high - temperature environment - </u>

In order to ionize a gas , i.e. to pull out an electron from a neutral atom , to form an ion .

The process of ionization require high amount of energy to remove the electron from the outer most shell of the atom .

The amount of energy for the process of ionization is only possible in high - temperature regions .

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Two children of mass 27 kg and 51 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the c
Flauer [41]

Answer:

2.62 m

Explanation:

Let the small child sit at a distance x from the pivot.

The distance of big child from the pivot is 4 - x .

By using the concept of moments.

Clockwise moments = anticlockwise moments

27 x = 51 ( 4 - x )

27 x = 204 - 51 x

78 x = 204

x = 2.62 m

8 0
3 years ago
What are sound waves
EastWind [94]

Answer:  Sound waves are a waves of compression and rarefaction, by which sound is propagated in an elastic medium such as air.

Explanation:

In physics, sound is a vibration that propagates as an acoustic wave, through a transmission medium such as a gas, liquid or solid.

8 0
3 years ago
Read 2 more answers
A balloon contains 0.075 m^3 of
inessss [21]

Answer:

600 KPa.

Explanation:

From the question given above, the following data were obtained:

Initial volume (V1) = 0.075 m³

Final volume (V2) = 0.45 m³

Final pressure (P2) = 100 KPa

Initial pressure (P1) =?

Temperature = constant

The initial pressure can be obtained by using the Boyle's law equation as shown below:

P1V1 = P2V2

P1 × 0.075 = 100 × 0.45

P1 × 0.075 = 45

Divide both side by 0.075

P1 = 45 / 0.075

P1 = 600 KPa.

Thus, the initial pressure in the balloon is 600 KPa.

7 0
2 years ago
A tennis ball is tossed upwards into the air with an initial velocity of +5m/s, how much time does it take for the tennis ball t
Sever21 [200]
HOPE THIS HELPS YOU!!!!!

3 0
2 years ago
Note: The rope is 20 m long. Answer like this: (1.<br> 2._____ etc)
qaws [65]

Answer:

1. <u>Potential energy</u>, 2. <u>Potential and kinetic energy</u>, 3. <u>Potential and kinetic energy</u>, 4. <u>Potential and kinetic energy</u>, 5. <u>Potential energy</u>

Explanation:

We note that the total mechanical energy (M.E.) of the body is given as follows;

M.E. = K.E. + P.E. = Constant

Where;

K.E. = The kinetic energy of the body = (1/2)·m·v²

P.E. = The potential energy of the body = m·g·h

m = The mass of the person

v = The velocity with which the person is in motion

g = The acceleration due to gravity ≈ 9.81 m/s²

h = The height of the person above the ground

The length of the rope = 20 m

The initial height at location 1, h₁ = 40.0 m

At location 1, the velocity, v₁ = 0.00 m/s

The mechanical energy, M.E. = K.E.₁ + P.E.₁

∴  K.E.₁ = 0 and P.E.₁ = m ×9.81×40

M.E. = (1/2) ×m ×0² + m ×9.81×40

∴ M.E. = 0 + P.E.₁ the type of energy present at location 1 is only potential energy

At location 2, the velocity, v₂ = 10.0 m/s

The mechanical energy, M.E. = K.E.₂ + P.E.₂ = (1/2) ×m ×10² + m ×9.81×40

∴  K.E.₂ = 50·m and P.E.₂ = m ×9.81×35 = 343.35·m

M.E. = 50·m + 343.35·m the type of energy at location 2 is both kinetic energy, K.E. and potential energy, P.E.

At location 3, the velocity, v₃ = 20.0 m/s

The mechanical energy, M.E. = K.E.₃ + P.E.₃ = (1/2) ×m ×20² + m ×9.81×20

∴  K.E.₃ = 200·m and P.E.₃ = m ×9.81×20 = 196.2·m

M.E. = 200·m + 196.2·m the type of energy at location 3 is both kinetic energy, K.E. and potential energy, P.E.

At location 4, the velocity, v₄² = 350.0 m²/s², h₄ = 15.0 m

The mechanical energy, M.E. = K.E.₄ + P.E.₄ = (1/2) × m ×350 + m ×9.81×15

∴  K.E.₄ = 175·m and P.E.₄ = m×9.81×15 = 147.15·m

M.E. = 175·m + 147.15·m the type of energy at location 4 is both kinetic energy, K.E. and potential energy, P.E.

At location 5, the velocity, v₅ = 0 m/s, h₅ = 10.0 m

The mechanical energy, M.E. = K.E.₅ + P.E.₅ = (1/2) × m × 0 + m ×9.81×10

∴  K.E.₅ = 0·m and P.E.₅ = m×98.1 = 98.1·m

M.E. = 0·m + 98.1·m the type of energy at location 5 is only potential energy, P.E.

Therefore, we have;

\left|\begin{array}{ccc}Location&&Type(s) \ of \ Energy \ Presents\\1&&Potential \ Energy\\2&&Potential  \ and \ Kinetic \ Energy\\3&&Potential  \ and \ Kinetic \ Energy\\4&&Potential  \ and \ Kinetic \ Energy\\5&&Potential  \  Energy\end{array} \right |

5 0
2 years ago
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