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velikii [3]
3 years ago
7

If the distance between 2 charged particles is double, the force between them changes by a factor of?

Physics
1 answer:
aev [14]3 years ago
3 0

Answer:

See explanation

Explanation:

Given that;

q1 and q2 are the magnitudes of the two charges while r1 is the distance between the charges.

F1 =kq1q2/r1^2

Then,

F2 = kq1q2/(2r1)^2

F2 = kq1q2/4r1^2

So,

F2 = 1/4 F1

The force between the charges changes by a factor of 1/4.

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A family is skating at an ice rink. The 58.2 kg mother is holding the
MariettaO [177]

Answer:

When I got this question I had to draw it out so if you have to do that, draw 3 stick figures holding hands, one representing the mother, father, and daughter. Then you write their weights on top of them and then draw an arrow pointing from the father to the mother.

Explanation:

use this formula :

a_{y} = \frac{Fdadshandy}{msys}

then you fill it in :

a_{y} = \frac{100N}{35.5kg+58.2kg}

a_{y} = \frac{100N}{93.7kg}

a_{y} = 1.0672 m/s^{2}

then you multiply that with the daughters weight :

T_{2} x= m_{2} a_{y}

T_{2} x = 35.5kg (1.0672 m/s^{2})

T_{2} x = 37.89N

and that's the answer :) : 37.89N

5 0
3 years ago
The following are examples of physical properties except
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6 0
3 years ago
Read 2 more answers
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Sonja [21]
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The answer is C 39 hz

Please look at the attached image below for the explanation

5 0
3 years ago
Vectors are an important part of the language of science, mathematics, and engineering.
tensa zangetsu [6.8K]

Answer:

<u><em>a. True</em></u>

Explanation:

<em>Vectors are an important part of the language of science, mathematics, and engineering.</em>

4 0
2 years ago
Light of wavelength 559 nm is used to illuminate normally two glass plates 22.1 cm in length that touch at one end and are separ
umka21 [38]

Answer:

M = 222 fringes

Explanation:

given

λ = 559 n m = 559 × 10⁻⁹ m

radius = 0.026 mm = 0.026 ×10⁻³ m

length of the glass plate = 22.1 ×10⁻² m

using relation

2t=(m+\dfrac{1}{2})\lambda\ \ (m=0,1,2,3...)\\where\ 0\leq t\leq 2r\\m = \dfrac{2t}{\lambda}-\dfrac{1}{2}

m_{max} = \dfrac{2\times 2r}{\lambda}-\dfrac{1}{2}\\m_{max} = \dfrac{2\times 2\times 0.026\times 10^{-3}}{559\times 10^{-9}}-\dfrac{1}{2}

 = 221.79  

 = 221 (approx.)

hence no of bright fringe

M = m + 1

   = 221 +1

M = 222 fringes

6 0
3 years ago
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