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kherson [118]
3 years ago
9

What causes an object to be attracted to another object due to electrostatic forces?​

Physics
1 answer:
oksano4ka [1.4K]3 years ago
5 0

Answer:...............

Explanation:

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What two quantities must stay the same in order for an object to have a constant velocity? A) The speed and kinetic energy must
GarryVolchara [31]

the answer is c) the speed and direction of travel must be constant


7 0
3 years ago
Points A, B, and C form the vertices of a triangle in a nonuniform electrostatic field. The electrostatic work done on a particl
Trava [24]

Answer:

Explanation:

Let electric potential at A ,B and C be Va , Vb and Vc respectively.

Work done = charge x potential difference

Wab = q ( Va - Vb )

Wac =  q (  Va -  Vc )

Given

Wac = - Wab / 3

3Wac = - Wab

Now

Wbc = q ( Vb - Vc )

= q [ ( Va-Vc ) - ( Va - Vb )]  

= Wac - Wab

= Wac + 3Wac

= 4Wac

4 0
3 years ago
Muscles make up ore than __ of a person body?<br>a.10<br>b.25<br>c40<br>d.75
exis [7]
40 percent of your body
8 0
3 years ago
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A flea jumps straight up to a maximum height of 0.550 m . what is its initial velocity v0 as it leaves the ground?
Alexxx [7]
Since my givens are x = .550m [Vsub0] = unknown
 [Asubx] = =9.80
 
 [Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]

[Vsubx]^2 = [Vsub0x]^2 + 2[Asubx] * (X-[Xsub0]) 

Vsubx is the final velocity, which at the max height is 0, and Xsub0 is just 0 as that's where it starts so I just plug the rest in

0^2 = [Vsub0x]^2 + 2[-9.80]*(.550)

0 = [Vsub0x]^2 -10.78

10.78 = [Vsub0x]^2

Sqrt(10.78) = 3.28 m/s 


3 0
3 years ago
When a cube is inscribed in a sphere of radius r, the length Lof a side of the cube is . If a positive point charge Qis placed a
Nana76 [90]

Answer:

  Ф_cube /Ф_sphere = 3 /π

Explanation:

The electrical flow is

      Ф = E A

where E is the electric field and A is the surface area

Let's shut down the electric field with Gauss's law

       Фi = ∫ E .dA = q_{int} / ε₀

the Gaussian surface is a sphere so its area is

        A = 4 π r²

the charge inside is

        q_{int} = Q

we substitute

       E 4π r² = Q /ε₀

       E = 1 / 4πε₀   Q / r²

To calculate the flow on the two surfaces

* Sphere

       Ф = E A

        Ф = 1 / 4πε₀  Q / r² (4π r²)

        Ф_sphere = Q /ε₀

* Cube

Let's find the side value of the cube inscribed inside the sphere.

In this case the radius of the sphere is half the diagonal of the cube

          r = d / 2

We look for the diagonal with the Pythagorean theorem

         d² = L² + L² = 2 L²

         d = √2 L

         

we substitute

          r = √2 / 2 L

          r = L / √2

          L = √2  r

now we can calculate the area of ​​the cube that has 6 faces

          A = 6 L²

          A = 6 (√2  r)²

          A = 12 r²

the flow is

          Ф = E A

          Ф = 1 / 4πε₀  Q/r²  (12r²)

          Ф_cubo = 3 /πε₀  Q

the relationship of these two flows is

         Ф_cube /Ф_sphere = 3 /π

8 0
3 years ago
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