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lana66690 [7]
3 years ago
14

Which of the following is a false statement about dispersion forces? View Available Hint(s) Which of the following is a false st

atement about dispersion forces? Dispersion forces are the result of fluctuations in the electron distribution within molecules or atoms. Dispersion forces are present in all atoms and molecules. Dispersion forces result from the formation of instantaneous dipoles in a molecule or atom. Dispersion forces always have a greater magnitude in molecules with a greater molar mass. Dispersion force magnitude depends on the amount of surface area available for interactions.
Physics
1 answer:
icang [17]3 years ago
4 0

Explanation :

Dispersion forces are also known as London dispersion forces. It is the weakest force. Also, it is the part of the Van der Waals forces.

(1) This force is exhibited by all atoms and molecules.

(2) These forces are the result of the fluctuations in the electron distribution within molecules or atoms. Due to these fluctuations, the electric field is created. The magnitude of this force is explained in terms of Hamaker constant 'A'.

(3) Dispersion forces result from the formation of instantaneous dipoles in a molecule or atom. When electrons are more concentrated in a place, instantaneous dipoles formed.

(4) Dispersion force magnitude depends on the amount of surface area available for interactions. If the area increases, the size of the atom also increase. As a result, stronger dispersion forces.

So, the false statement is "Dispersion forces always have a greater magnitude in molecules with a greater molar mass".  

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REY [17]

Answer:

it's A, D, E

Explanation:

6 0
3 years ago
Read 2 more answers
Refer to the first diagram. What is the weight of the person hanging on the end of the seesaw in Newtons?
irina1246 [14]

Due to equilibrium of moments:

1) The weight of the person hanging on the left is 250 N

2) The 400 N person is 3 m from the fulcrum

3) The weight of the board is 200 N

Explanation:

1)

To solve the problem, we use the principle of equilibrium of moments.

In fact, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

The moment of a force is defined as:

M=Fd

where

F is the magnitude of the force

d is the perpendicular distance of the force from the fulcrum

In the first diagram:

- The clockwise moment is due to the person on the right is

M_c = W_2 d_2

where W_2 = 500 N is the weight of the person and d_2 = 2 m is its distance from the fulcrum

- The anticlockwise moment due to the person hanging on the left is

M_a = W_1 d_1

where W_1 is his weight and d_1 = 4 m is the distance from the fulcrum

Since the seesaw is in equilibrium,

M_c = M_a

So we can find the weight of the person on the left:

W_1 d_1 = W_2 d_2\\W_1 = \frac{W_2 d_2}{d_1}=\frac{(500)(2)}{4}=250 N

2)

Again, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment due to the person on the right is

M_c = W_2 d_2

where W_2 = 400 N is the weight of the person and d_2 is its distance from the fulcrum

- The anticlockwise moment due to the person on the left is

M_a = W_1 d_1

where W_1 = 300 N is his weight and d_1 = 4 m is the distance from the fulcrum.

Since the seesaw is in equilibrium,

M_c = M_a

So we can find the distance of the person on the right:

W_1 d_1 = W_2 d_2\\d_2 = \frac{W_1 d_1}{W_2}=\frac{(300)(4)}{400}=3 m

3)

As before, for the seesaw to be in equilibrium, the total clockwise moment must be equal to the total anticlockwise moment.

- The clockwise moment around the fulcrum this time is due to the weight of the seesaw:

M_c = W_2 d_2

where W_2 is the weight of the seesaw and d_2 = 3 m is the distance of its centre of mass from the fulcrum

- The anticlockwise moment due to the person on the left is

M_a = W_1 d_1

where W_1 = 600 N is his weight and d_1 = 1 m is the distance from the fulcrum

Since the seesaw is in equilibrium,

M_c = M_a

So we can find the weight of the seesaw:

W_1 d_1 = W_2 d_2\\W_2 =\frac{W_1 d_1}{d_2}= \frac{(600)(1)}{3}=200 N

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8 0
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Write the law of conservation of energy. How does it apply to eating and exercising?
zzz [600]
The Law of Conservation of energy states that the total energy of an isolated system remains constant-it is said to be the conserved over time. It applies by us burning calories because humans can lose or gain weight.
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by scale drawing, find the resultant of vectors 70N inclined at100N and direction of the resultant 100N​
Thepotemich [5.8K]

Answer: 70n=

<h3>at100n=</h3>

100n=

Explanation:

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Alexxandr [17]

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Explanation:

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where

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v is its final speed

u is its initial speed

In this case, the force does a positive amount of work on the object, so

W>0

This also implies that

K_f > K_i

And so

\frac{1}{2}mv^2 > \frac{1}{2}mu^2

And therefore

v>u

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