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NeTakaya
3 years ago
7

A small sphere with mass 5.00.×10−7kg and charge +3.00μC is released from rest a distance of 0.400 m above a large horizontal in

sulating sheet of charge that has uniform surface charge density σ=+8.00pC/m2.
Physics
1 answer:
julia-pushkina [17]3 years ago
6 0
I will assume you are asking what the initial acceleration of the sphere is since the information provided seems to indicate that.
First we need to know Newton's Law
F=ma.  
We know the mass of the sphere and we want a so we solve to get
a=F/m.
Now we need the force on the charged sphere.  This is given by the electric field, E and the charge, Q.  The relationship is F=Q×E.  (Recall that the electric field units can be expressed in Newtons/Coulomb).
Now the electric field above a large (~infinite) sheet of charge with a known charge density σ, is given by
E = σ/(2ε0)
Plug in your values of σ, to get E, then the sphere charge Q to get F, the the mass into a = F/m to get the acceleration

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Read 2 more answers
Assuming that each nucleus is roughlyspherical and that its mass is roughly equal to A (in atomic mass units {\rm u}), what is t
lara [203]

Answer:

ρ/ρ2 = 3 / R₀       the two densities are different

Explanation:

Density is defined as

       ρ = M / V

As the nucleus is spherical

       V = 4/3 π r³

Let's replace

      ρ = A / (4/3 π R₀³)

      ρ = ¾ A / π R₀³

b)

      ρ2 = F / area

The area of ​​a sphere is

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     ρ2 = F / 4π R₀²

     ρ2 = F / 4π R₀²

Atomic number is the number of protons in the nucleon in not very heavy nuclei. This number is equal to the number of neutrons, but changes in heavier nuclei, there are more neutrons than protons.

Let's look for the relationship of the two densities

     ρ/ρ2 = ¾ A / π R₀³ / (F / 4π R₀²)

     ρ /ρ2 = 3 (A / F) (1 / R₀)

In this case it does not say that the nucleon number is A (F = A), the relationship is

     ρ/ρ2 = 3 / R₀

I see that the two densities are different

3 0
3 years ago
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