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What three things are carried throughout the body by the cardiovascular system

Physics

1 answer:

Natasha_Volkova [10]3 years ago

8 0

Blood, vitamins, and cells :)

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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

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

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