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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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A stone is catapulted at time t = 0, with an initial velocity of magnitude 19.9 m/s and at an angle of 39.9° above the horizonta
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Answer:

Part a)

x = 15.76 m

Part b)

y = 7.94 m

Part c)

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Part d)

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Part e)

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Part f)

y = -75.6 m

Explanation:

As we know that catapult is projected with speed 19.9 m/s

so here we have

v_x = 19.9 cos39.9

v_x = 15.3 m/s

similarly we have

v_y = 19.9 sin39.9

v_y = 12.76 m/s

Part a)

Horizontal displacement in 1.03 s

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x = (15.3)(1.03)

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Part b)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.03) - 4.9(1.03)^2

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Part c)

Horizontal displacement in 1.71 s

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x = (15.3)(1.71)

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Part d)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

y = (12.76)(1.71) - 4.9(1.71)^2

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Part e)

Horizontal displacement in 5.44 s

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x = 83.23 m

Part f)

Vertical direction we have

y = v_y t - \frac{1]{2}gt^2

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3 years ago
A block of mass 0.221 kg is placed on top of a light, vertical spring of force constant 5365 N/m and pushed downward so that the
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Answer:

The maximum height above the point of release is 11.653 m.

Explanation:

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Mass of block = 0.221 kg

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Using stored energy in spring

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Using gravitational potential energy

U' =mgh....(II)

Using energy of conservation

E_{i}=E_{f}

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