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Tasya [4]
3 years ago
14

Which statement bet explains the relationship between the electric force between two charged objects and the distance between th

em
Physics
1 answer:
Nataliya [291]3 years ago
4 0
Coulomb's Law: Force = k x q1x q2 divided distance square
where k=9x10^9 , q1 and q2 are the charge
So if you distance is halved, your force is stronger by 4 times
and if you distance is doubled, your force is 1/4
Ask me again if you aren't clear :)

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3 years ago
S Problem Set<br> 2.) 6.4 x 109 nm to cm
anyanavicka [17]

Answer:

6.4\cdot 10^2 cm

Explanation:

First of all, let's convert from nanometres to metres, keeping in mind that

1 nm = 10^{-9} m

So we have:

6.4\cdot 10^9 nm \cdot 10^{-9} m/nm = 6.4 m

Now we can convert from metres to centimetres, keeping in mind that

1 m = 10^2 cm

So, we find:

6.4 m \cdot 10^2 cm/m = 6.4\cdot 10^2 cm

8 0
3 years ago
The free-body diagrams of four objects are shown.
kipiarov [429]

Answer:

its unbalanced and balanced

Explanation:

5 0
3 years ago
Two parallel plates are a distance apart with a potential difference between them. A point charge moves from the negatively char
IgorC [24]

Answer:

K' = 1200 J

Explanation:

To find the kinetic energy you first take into account the formula for the kinetic energy of the charge:

K=\frac{1}{2}mv^2 = 800J   (1)

m: mass of the charge

v: final speed of the charge when it reaches the positively charged plate.

Furthermore, you have that the acceleration of the charge is obtained by using the second Newton law:

F=ma=qE\\\\a=\frac{qE}{m} (2)

a: acceleration

E: electric field

q: charge

The electric field between two parallel plates is V/d, being V the potential difference and d the separation between plates. You replace E in (2) and obtain:

a=\frac{qV}{md}

Next, you take into account the following formula for the calculation of the final speed of the charge:

v^2=v_o^2+2ad\\\\v_o=0m/s\\\\v=\sqrt{\frac{2qVd}{md}}=\sqrt{\frac{2qV}{m}}

Next, you replace this value of v in (1):

K=\frac{1}{2}mv^2=\frac{1}{2}m(\frac{2qV}{m})=qV = 880J   (3)

If the distance between plates is tripled, and the potential difference is halved, you have for the new final speed:

v'^2=v'_o^2+2a(3d)\\\\v_o=0m/s\\\\v'=\sqrt{6ad}=\sqrt{6(\frac{q}{md})\frac{V}{2}d}=\sqrt{\frac{3qV}{m}}

And the kinetic energy becomes:

K'=\frac{1}{2}mv^2=\frac{1}{2}m(\frac{3qV}{m})=\frac{3}{2}qV    (4)

You calculate the ratio between both kinetic energies K and K', that is, you divide equations (3) and (4), in order to find the new kinetic energy:

K=qV=800J\\\\K'=\frac{3}{2}qV\\\\\frac{K}{K'}=\frac{qV}{3/2\ qV}=\frac{2}{3}\\\\K'=\frac{3}{2}K=\frac{3}{2}(800J)=1200J

hence, the kinetic energy of the charge incresases to 1200J

4 0
3 years ago
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