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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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Calculate the wavelength λ1 for gamma rays of frequency f1 = 7.20×1021 hz .
bagirrra123 [75]
The relationship between wavelength \lambda, frequency f and speed of light c for an electromagnetic wave is
\lambda= \frac{c}{f}
Using the data of the problem, we find
\lambda= \frac{3\cdot 10^8 m/s }{7.20 \cdot 10^{21} Hz}=4.17 \cdot 10^{-14} m
5 0
3 years ago
Light moves at a speed of around 1 million miles per hour<br>O<br>True<br>False​
AURORKA [14]

Answer:

False

Explanation:

In miles per hour, light speed is about 670,616,629 mph

5 0
3 years ago
Read 2 more answers
The true brightness of an object is called its
g100num [7]
<span>The true brightness of an object is called its luminosity. It is the total amount of energy emitted by bright or meteorological objects over a period of time. It has the SI unit of joules per second or watts. So the answer is letter A. Intensity is the measure of how strong the substance or object is when it projects something. Magnitude is a measure of how great is the size the object produces. Viscosity is the measure of flow of a substance.</span>
8 0
3 years ago
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Water waves in a small tank are 6.6 m long. They travel at a speed of 2.9 m/s. What is the frequency of the water
liubo4ka [24]

Answer:

.439 Hz

Explanation:

Wavelength = 6.6 m

speed = 2.9 m/s

  2.9 = 6.6 f

f = 2.9/6.6 =  .439 Hz

8 0
1 year ago
Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

8 0
3 years ago
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