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pychu [463]
3 years ago
11

The energy required to remove an electron from a gaseous atom is called

Physics
1 answer:
IRISSAK [1]3 years ago
6 0

Answer: The energy required to remove an electron from a gaseous atom is called ionization energy.

Explanation:

Ionization energy is defined as the energy required to remove the most loosely bound electron from a neutral gaseous atom.

When we move across a period from left to right then there occurs a decrease in atomic size of the atoms. Therefore, ionization energy increases along a period but decreases along a group.

Smaller is the size of an atom more will be the force of attraction between its protons and electrons. Hence, more amount of energy is required to remove an electron.

Thus, we can conclude that the energy required to remove an electron from a gaseous atom is called ionization energy.

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When two systems in contact are not at the same temperature, _____ occurs.expansionheat flowfrictionevaporation
Murljashka [212]

Heat flow occurs when two systems in contact are not at the same temperature.

What is heat flow?

  • when two bodies of different temperatures are in contact, heat flow takes place.
  • Heat flows till the two bodies in contact achieve equilibrium.
  • To achieve equilibrium, heat flows from hotter bodies to colder bodies.
  • There is no heat flow after achieving the state of equilibrium because the amount of heat flow from one body to the other is the same.

For more information, please visit: brainly.com/question/11297584?referrer=searchResults

#SPJ4

8 0
2 years ago
A regulation basketball has a 15 cm diameter and may be appropriated as a thin spherical shell.
REY [17]

Solution :

From the given data,

For the spherical shell is $\frac{k^2}{R^2}= \frac{2}{3}$

where x is the radius of gyration and the acceleration of a rolling body on an inclined plane = a

Therefore,

$a =\frac{g \sin \theta}{1+ \frac{k^2}{R^2}} $

  $= \frac{g \sin \theta}{1 +\frac{2}{3}}$

  $= \frac{3}{5} g \sin \theta$

  = 0.6 x 9.81 x sin ( 29.7)

  $= 2.913 \ m/s^2$

$h = \frac{1}{2} a(\Delta t)^2$

$\Delta t = \sqrt{\frac{2h}{a}}$

$\Delta t = \sqrt{\frac{2 \times 2.9}{2.913}}$

Δt = 1.411 s

6 0
3 years ago
If planet A is three times as far from planet C, then the period of its orbit will be __ times as long
liubo4ka [24]
I may be wrong, but I think you're trying to say that Planet-A is
<em>3 times as far from the sun</em> as Planet-C is.

If that's the real question, then the answer is that the period of Orbit-A
is about<em>  5.2</em>  times as long as the period of Orbit-C .

Orbital period ≈ (proportional to) (the orbital distance) ^ 3/2 power.

This was empirically demonstrated about 350 years ago by Johannes
and his brilliant Kepple, and derived about 100 years later by Newton
from his formula for the forces of gravity.


6 0
3 years ago
A ball of mass m is thrown straight upward from ground level at speed v0. At the same instant, at a distance D above the ground,
Alla [95]

Answer:

Explanation:

Let the balls collide after time t .

distance covered by falling ball

s₁ = v₀ t + 1/2 g t²

distance covered by rising ball

s₂ = v₀ t - 1/2 g t²

Given ,

s₁ + s₂ = D

D = v₀ t + 1/2 g t² + v₀ t - 1/2 g t²

= 2v₀ t

t = D / 2v₀

s₂ = v₀ t - 1/2 g t²

= v₀ x D / 2v₀ - (1/2) x  g x D² / 4v₀²

= D / 2 - gD² / 8 v₀²

6 0
3 years ago
two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at 20ms^-1 and the seco
Reika [66]

Answer:

t=750s

Explanation:

The two cars are under an uniform linear motion. So, the distance traveled by them is given by:

\Delta x=vt\\x_f-x_0=vt\\x_f=x_0+vt

x_f is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:

x_0_1=6km\\x_0_2=0

We have x_f_1=x_f_2. Thus, solving for t:

x_0_1+v_1t=x_0_2+v_2t\\x_0_1=t(v_2-v_1)\\t=\frac{x_0_1}{v_2-v_1}\\t=\frac{6*10^3m}{28\frac{m}{s}-20\frac{m}{s}}\\t=750s

8 0
3 years ago
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