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ANTONII [103]
3 years ago
8

Three identical point charges ( 2.0 nc are placed at the corners of an equilateral triangle with sides of 2.0-m length. if the e

lectric potential is taken to be zero at infinity, what is the potential (in v at the midpoint of any one of the sides of the triangle?
Physics
1 answer:
WARRIOR [948]3 years ago
3 0
Try the distance formula if you have the coordinates of the triangle (point to point) make sure they are congruent 
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Which circuit is a series circuit
posledela

Answer:

The answer is A

Explanation:

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49. \ A rectangular plate is rotating with a constant angular speed about an axis that passes perpendicularly through one corner
faust18 [17]

Answer:

    \frac{L_1}{L_2} = \sqrt{(n^2 - 1)}

Explanation:

For this interesting problem, we use the definition of centripetal acceleration  

      a = v² / r  

angular and linear velocity are related  

     v = w r  

we substitute  

    a = w² r

the rectangular body rotates at an angular velocity w  

We locate the points, unfortunately the diagram is not shown. In this case we have the axis of rotation in a corner, called O, in one of the adjacent corners we call it A and the opposite corner A  

the distance OB = L₂  

the distance AB = L₁

the sides of the rectangle  

It is indicated that the acceleration in in A and B are related  

      a_A = n \ a_B  

we substitute the value of the acceleration  

    w² r_A = n r_B  

the distance from the each corner is  

    r_B = L₂  

    r_A = \sqrt{L_1^2 + L_2^2}  

we substitute  

   \sqrt{L_1^2 + L_2^2} = n L₂  

    L₁² + L₂² = n² L₂²  

    L₁² = (n²-1) L₂²  

4 0
3 years ago
If 2 objects are moved by the same force (F):
kipiarov [429]

Answer:y=mx+b 58+5

Explanation:

8 0
3 years ago
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What is the gravitational force on a 70kg that is 6.38x10^6m above the earths surface
NARA [144]

Answer:

171.5 N

Explanation:

The gravitational force on an object due to the Earth is given by

F=mg

where

m is the mass of the object

g is the acceleration due to gravity

The acceleration due to gravity at a certain height h above the Earth is given by

g=\frac{GM}{(R+h)^2}

where:

G is the gravitational constant

M=5.98\cdot 10^{24} kg is the Earth's mass

R=6.37\cdot 10^6 m is the Earth's radius

Here,

h=6.38\cdot 10^6  m

So the acceleration due to gravity is

g=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})}{(6.37\cdot 10^6 + 6.38\cdot 10^6)^2}=2.45 m/s^2

We know that the mass of the object is

m = 70 kg

So, the gravitational force on it is

F=mg=(70)(2.45)=171.5 N

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3 years ago
the refractive index is the blank of the speed of light in a medium to the speed of light in a vacuum
andrew11 [14]

Answer:ratio

Explanation:

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