Question 2 of 20

Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance

SSSSS [86.1K]

Answer:

${F_{tot} = -1.092*10^{-2}N$

Explanation:

The question: What is the net force exerted by these two charges on a third charge q_3 = 47.0 nC placed between q_1 and q_2 at x_3 = -1.240 mm ?

<u>Your answer may be positive or negative, depending on the direction of the force.</u>

Solution:

The coulomb force is given by the equation

$F =k\dfrac{q_1 q_2}{d^2}.$

where $d$ is the separation between the charges $q_1$ and $q_2$ .

Now, in our case

$q_1=-13.5nC=-13.5*10^{-9}C$

$q_2=-1.735nC=-1.735*10^{-9}C$

$q_3= 47*10^{-9}C$

The separation between charges $q_3$ and $q_1$ is

$d_{31}= (-1.240mm)-(-1.735mm)=0.495mm=4.95*10^{-4} m$

Therefore, the force between them is

$F_{31} = (9*10^9NmC^{-2})\dfrac{47*10^{-9}*(-13.5*10^{-9})}{4.95*10^{-4}} =-0.01152N$

and it is directed in the negative x-direction.

The separation between charges $q_2$ and $q_3$ is

$d_{23} =-1.240mm-0=-1.240*10^{-3}m$

therefore, the force between them is

$F_{23} =(9*10^9Nm^2C^{-2})\dfrac{47*10^{-9}C*(-1.735*10^{-9}C)}{-1.240*10^{-3}m}=5.94*10^{-4}N$

Therefore the total force on charge $q_3$ is

$F_{tot} =5.94*10^{-4}-0.01152N = -0.010926N\\\\\boxed{F_{tot} = -1.092*10^{-2}N}$

8 0

3 years ago

Observe yourself breathing and count the number of times you inhale per second. During each breath you probably inhale 0.66 L of

Pavel [41]

To solve this exercise it is necessary to apply the concepts related to Robert Boyle's law where:

$PV=nRT$

Where,

P = Pressure

V = Volume

T = Temperature

n = amount of substance

R = Ideal gas constant

We start by calculating the volume of inhaled O_2 for it:

$V = 21\% * 0.66L$

$V = 0.1386L$

Our values are given as

P = 1atm

T=293K $R = 0.083145kJ*mol^{-1}K^{-1}$

Using the equation to find n, we have:

$PV=nRT$

$n = \frac{PV}{RT}$

$n = \frac{(1)(0.1386)}{(0.0821)(293)}$

$n = 5.761*10^{-3}mol$

Number of molecules would be found through Avogadro number, then

$\#Molecules = 5.761*10^{-3}*6.022*10^{23}$

$\#Molecules = 3.469*10^{21} molecules$

7 0

3 years ago

According to Newton’s first law of motion, when will an object at rest begin to move?

Natasha_Volkova [10]

Newton's first law of motion says something like "An object remains
in constant, uniform motion until acted on by an external force".

Constant uniform motion means no change in speed or direction.
If an object changes from rest to motion, that's definitely a change
of speed. So it doesn't remain in the state of constant uniform
motion (none) that it had when it was at rest, and that tells us
that an external force must have acted on it.

8 0

4 years ago

Read 2 more answers

A certain radioactive nuclide decays with a disintegration constant of 0.0178 h-1.

umka21 [38]

Explanation:

Given that,

The disintegration constant of the nuclide, $\lambda=0.0178\ h^{-1}$

(a) The half life of this nuclide is given by :

$t_{1/2}=\dfrac{ln(2)}{\lambda}$

$t_{1/2}=\dfrac{ln(2)}{0.0178}$

$t_{1/2}=38.94\ h$

(b) The decay equation of any radioactive nuclide is given by :

$N=N_oe^{-\lambda t}$

$\dfrac{N}{N_o}=e^{-\lambda t}$

Number of remaining sample in 4.44 half lives is :

$t_{1/2}=4.44\times 38.94$

$t_{1/2}=172.89\ h^{-1}$

So, $\dfrac{N}{N_o}=e^{-0.0178\times 172.89}$

$\dfrac{N}{N_o}=0.046$

(c) Number of remaining sample in 14.6 days is :

$t_{1/2}=14.6\times 24$

$t_{1/2}=350.4\ h^{-1}$

So, $\dfrac{N}{N_o}=e^{-0.0178\times 350.4}$

$\dfrac{N}{N_o}=0.0019$

Hence, this is the required solution.

4 0

3 years ago

Describe the difference between a scientific law and a scientific theory

Luda [366]

A scientific law is absolute, it cannot be proven wrong. A scientific theory is like a belief, it has been proven but can also be argued or disproven.

3 0

4 years ago

Read 2 more answers