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Dovator [93]
3 years ago
10

Pause

Physics
1 answer:
nirvana33 [79]3 years ago
3 0

Answer:

C

Explanation:

a=f/m

10Kgm/s2/4kg

2.5m/s2

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Which of the following is a false statement about dispersion forces? View Available Hint(s) Which of the following is a false st
icang [17]

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".  

4 0
3 years ago
A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o
Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field dE at the center of curvature of the arc is

dE = k\dfrac{dQ}{r^2}cos(\theta ) <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where r is the radius of curvature.

Now

dQ = \lambda rd\theta,

where \lambda is the charge per unit length, and it has the value

\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.

Thus, the electric field at the center of the curvature of the arc is:

E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta  }{r^2} cos(\theta)

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.

Now, we find \theta_1 and \theta_2. To do this we ask ourselves what fraction is the arc length  3.0 of the circumference of the circle:

fraction = \dfrac{3.0m}{2\pi (2.3m)}  = 0.2076

and this is  

0.2076*2\pi =1.304 radians.

Therefore,

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.

evaluating the integral, and putting in the numerical values  we get:

E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\

\boxed{ E = 31.329N/C.}

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3 years ago
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3 years ago
Black holes must have a mass three or more times the mass of _____.
tia_tia [17]
The answer is <span>d. the sun</span>
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3 years ago
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At the train station, you notice a large horizontal spring at the end of the track where the train comes in. This is a safety de
e-lub [12.9K]

Assume a maximum stopping acceleration of g/2 where g is acceleration due to gravity.

Answer:

2.99 m/s

Explanation:

Stopping distance, s = 3 ft = 0.914 m

final velocity, v = 0

a = g/2 = 4.9 m/s²

Use third equation of motion:

v^2-u^2 = 2as

substitute the values to find the speed of train:

0 -u^2 = 2\times -4.9 \times 0.914 \\u^2=8.96 \\u=2.99 m/s

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3 years ago
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