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3 years ago

Which molecule will undergo only london dispersion forces when interacting with other molecules of the same kind?

Physics

1 answer:

vfiekz [6]3 years ago

7 0

Explanation:

London dispersion forces will form between non-polar molecules(polar ) that are symmetrical like O₂, H₂, Cl₂ and noble gases.

The attraction here is because non-polar molecules becomes polar due to the constant motion of its electrons.
This lead to an uneven charge distribution at an instant.
A temporary dipole or instantaneous dipole forms.
The temporary dipole can induce neighboring molecules to be distorted and forms dipoles as well.
This forms london dispersion forces.

Learn more:

Intermolecular forces brainly.com/question/10602513

#learnwithBrainly

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A brick and a feather fall to the earth at their respective terminal velocities. Which objectexperiences the greater force of ai

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Answer:

Under the reasonable assumption that the brick has more mass than the feather, the brick experiences a greater force of air friction.

Explanation:

Objects at terminal velocity, only under the influence of gravity, have maximized their speed and have an acceleration of zero. Thus, neither object is accelerating.

Recall Newton's second law: $\sum {\vec {F}}=m \vec {a}$

Since acceleration for each object is zero, the sum of the force acting on each of those objects must also be zero.

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Recall that near the surface of the earth, $F_{gravity}=mg$ , so the Force of Gravity acting on an object is directly proportional to the object's mass. (A similar argument could be made even if this were not taking place on the surface of the earth, so long as the objects were the same distance from the object providing gravitational influence).

If the masses of the objects are different, the object with the greater mass will experience a larger force of gravity, and hence a larger force of air friction at terminal velocity.

Under the reasonable assumption that the brick has more mass than the feather, the brick experiences a greater force of air friction.

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