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yanalaym [24]
3 years ago
5

Which of these was a characteristic of the Rutherford model of the atom, but not the Thomson model? (2 points)

Physics
2 answers:
Masteriza [31]3 years ago
8 0

answer B is the correct answer

tia_tia [17]3 years ago
3 0
Thomsons model( commonly the plum pudding model) - can also be described by a watermelon cut horizontally. The red of the water melon is a positively charged cloud with the seeds representing the negatively charged electrons sprinkled through out the positive cloud. 

Rutherford had a model with a dense positively charged nucleus at the center. This nucleus was encircled by negatively charged atoms.

Answer: <span>b. positive charge concentrated at the center of the atom</span>
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A vector in the xy plane has components -14.0 units in the x-direction and 30.0 units in the y-direction. What is the magnitude
OverLord2011 [107]

\huge\underline{\underline{\boxed{\mathbb {SOLUTION:}}}}

We would calculate the magnitude by applying pythagorean theorem:

\longrightarrow \sf{Magnitude= \sqrt{(-14)^2 } + 30^2}

\longrightarrow \sf{Magnitude = 33.12}

\longrightarrow \sf{The \:  vector \: is \: (- 14, 30)}

The angle between two vectors is given by the formula:

\sf{\longrightarrow \small  \cos \emptyset =  \dfrac{(a1b1 + a2b2)}{ \sqrt{(a1)^2 + (a2)^2√(b1)^2 + (b2)^2} } }

In two dimensional, the x axis of vector form is:

\small\sf{\longrightarrow (b1, b2) = (1, 0) }

\sf{\longrightarrow \small \cos \:  \emptyset =  \dfrac{(14 * 1 + 30 x 0)}{( \sqrt{(-14)^2 + (30)^2)(√(1)^2 + (0)^2)} } }

\small\longrightarrow \sf{ \dfrac{14}{33.12} }

\small\longrightarrow \sf{\emptyset  \:  = arcCos (\dfrac{ - 14}{33.12} )}

\small\longrightarrow \sf{\emptyset= 115^\circ}

\huge\underline{\underline{\boxed{\mathbb {ANSWER:}}}}

\small\bm{The \: angle  \: between  \: the  \: vector \: }

\small\bm{and \: \: the \: \: positive \: \: x \: \: axis \: \: is \: \: \: 115^\circ .}

6 0
2 years ago
Plz i need help for the 5 problems. plz show the work!!!
Artemon [7]

Answer:

1.   3 m/s^{2}

2.   1.5 m/s^{2}

3.   3 seconds

4.   0 m/s^{2}

5.   2.2 seconds

Explanation:

(1)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds

a = \frac {30-0}{10}=3 m/s^{2}

(2)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=10m/s, v=22m/s and t=8 seconds

a = \frac {22-10}{8}=1.5 m/s^{2}

(3)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=0m/s since at rest, v=15m/s and a=5 \frac {m}{s^{2}}

= \frac {15-0}{5}=3s

(4)

When initial and final velocity are constant, there’s no acceleration as proven below

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds

a = \frac {20-20}{10}=0 m/s^{2}

(5)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=9m/s since at rest, v=0m/s and a=-4.1 \frac {m}{s^{2}}

= \frac {0-9}{-4.1}=2.2s

8 0
3 years ago
Which equations could be used as is, or rearranged to calculate for frequency of a wave? Check all that apply.
amm1812
-- Equations  #2  and  #6  are both the same equation,
and are both correct.

-- If you divide each side by  'wavelength', you get Equation #4,
which is also correct.

-- If you divide each side by  'frequency', you get Equation #3,
which is also correct. 
With some work, you can rearrange this one and use it to calculate
frequency.

Summary:

-- Equations #2, #3, #4, and #6 are all correct statements,
and can be used to find frequency.

-- Equations #1 and #5 are incorrect statements.
7 0
3 years ago
Read 2 more answers
What are the relationships between position, velocity, and acceleration as it relates to time?
Ber [7]
Velocity = distance /time
acceleration = velocity / time
5 0
3 years ago
Read 2 more answers
A power supply is connected to a 59 Ohm resistor and a 53 Ohm resistor in series. The total current is found to be 0.15 A. What
kirill [66]

the answer is 47265.dug

Explanation:

im bot sure

4 0
2 years ago
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