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scoray [572]
3 years ago
11

How did rutherfords discovery effect the model of the atom

Physics
1 answer:
Kamila [148]3 years ago
7 0
<h2>Thomson proposed the plum pudding model of the atom, which had negatively-charged electrons embedded within a positively-charged "soup." Rutherford's gold foil experiment showed that the atom is mostly empty space with a tiny, dense, positively-charged nucleus.</h2>
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What would you have loved to press the pause button on so you could go deeper
kiruha [24]

Answer:

~Banana Fish~

3 0
3 years ago
three condensers are connected in series across a 150 volt supply. The voltages across them are 40,50 and 60 volts respectively,
ioda

Explanation:

Given that,

The voltages across them are 40,50 and 60 volts respectively, and the charge on each condenser is 6×10⁻⁸ C.

(a) Capacitance of capacitor 1,

C_1=\dfrac{Q}{V_1}\\\\C_1=\dfrac{6\times 10^{-8}}{40}\\\\C_1=1.5\times 10^{-9}\ F\\\\C_1=1.5\ nF

Capacitance of capacitor 2,

C_2=\dfrac{Q}{V_2}\\\\C_2=\dfrac{6\times 10^{-8}}{50}\\\\C_2=1.2\times 10^{-9}\ F\\\\C_2=1.2\ nF

Capacitance of capacitor 3,

C_3=\dfrac{Q}{V_3}\\\\C_3=\dfrac{6\times 10^{-8}}{60}\\\\C_3=1\times 10^{-9}\ F\\\\C_3=1\ nF

(b) The equivalent capacitance in series combination is :

\dfrac{1}{C}=\dfrac{1}{C_1}+\dfrac{1}{C_2}+\dfrac{1}{C_3}\\\\\dfrac{1}{C}=\dfrac{1}{1.5}+\dfrac{1}{1.2}+\dfrac{1}{1}\\\\C=0.4\ nF

Hence, this is the required solution.

5 0
3 years ago
The speed that light travels :
OLga [1]
I think it might be C.
4 0
3 years ago
Read 2 more answers
A thin spherical shell of radius R has a total charge +Q uniformly distributed over its surface. Of the following distance r fro
grigory [225]

Answer:

The correct answer is B

Explanation:

Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity

         Φ._{E} = ∫ E. dA = q_{int} / ε₀

For this case we create a Gaussian surface that is a sphere.  We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product

        ∫ E dA = q_{int} / ε₀

The area of ​​a sphere is

     A = 4π r²

   

    E 4π r² =q_{int} / ε₀

    E = (1 /4πε₀ )  q / r²

Having the solution of the problem let's analyze the points:

A   ) r = 3R / 4  = 0.75 R.

  In this case there is no charge inside the Gaussian surface therefore the electric field is zero

        E = 0

B) r = 5R / 4 = 1.25R

In this case the entire charge is inside the Gaussian surface, the field is

    E = (1 /4πε₀ )  Q / (1.25R)²

    E = (1 /4πε₀ )  Q / R2 1 / 1.56²

    E₀ = (1 /4π ε₀ )  Q / R²

   E_{B} =  Eo /1.56 ²

  E_{B}  = 0.41 Eo

C) r = 2R

All charge inside is inside the Gaussian surface

    E_{B} =(1 /4π ε₀ ) Q    1/(2R)²

    E_{B} = (1 /4π ε₀ ) q/R²   1/4

    E_{B} = Eo  1/4

    E_{B} = 0.25 Eo

D) False the field changes with distance

The correct answer is B

4 0
3 years ago
I need help asap pls
tigry1 [53]

no.9 indeed all matters are made out of atoms which is smaller than protons and neutrons

6 0
3 years ago
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