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2 years ago

How as bohr's atomic model similar to Rutherford's model

Physics

2 answers:

zysi [14]2 years ago

6 0

Answer:

It described a nucleus surrounded by a large volume of space

Explanation:

In 1911 Rutherford, based on the results of his famous gold foil experiment, proposed that the atoms are made of positively charged and highly denser nucleus, around which the negatively charged and weightless electrons are revolving.

But he failed to provide the reason why the electrons are not continuously emitting electromagnetic radiation, since according to classical physics charged particle in circular motion emits radiation.

in 1913 Bohr improved this model by the quantization of orbits.

According to his theory electrons are revolving in stable orbits at particular radius and they emit radiation only when the move from one orbit to another.

So description of a nucleus surrounded by a large volume of space is the similarity to Rutherford’s model.

Lemur [1.5K]2 years ago

5 0

It described a nucleus surrounded by a large volume of space. ... Electrons travel around the nucleus in fixed energy levels with energies that vary from level to level. C) Electrons travel around the nucleus in fixed energy levels with equal amounts of energy.

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BRAINLIEST AND MAY DOUBLE POINTS!!!

raketka [301]

Given that:

Energy of bulb (Work ) = 30 J,

Time (t) = 3 sec

The power consumption = ?

We know that, Power can be defined as rate of doing work

Power (P) = Work(Energy supplied) ÷ time

= 30 ÷ 3

= 10 Watts

<em> The power consumption is 10 W.</em>

3 0

3 years ago

A body weights 50 N in air and 45 N when wholly immersed in water calculate (i) the loss in weight of the body in water (ii) the

Lelechka [254]

Answer:

$difference \: in \: weight = 150n - 100n = 50n$

Now,buyantant force

$difference \: in \: weight \: = volume(body) \times density \: of \: water \: \times g$

so;

$50 = {v}^{b} \times 1 \times {10}^{3} \times 9.8m {s}^{2}$

${v}^{b} = \frac{50}{1000 } \times 9.8$

$= \frac{50}{9800}$

$= 0.0051$

Now,

$mass \: in \: air \: = 150n = \frac{150}{9.8kg}$

$density = \frac{weght}{volume}$

$= \frac{150}{0.0051} \times 9.8 \\ x = 3000$

And now,

$specific \: density \: = \frac{density of \: the \: body}{density \: of \: water}$

$= \frac{3000}{1000}$

$= 3$

Hence that,specific density of a given body is 3

please mark me as brainliest, please

3 0

2 years ago

Hope you all doing okay

Vsevolod [243]

Answer:Thank you

Explanation:

8 0

1 year ago

ly charged particles are held 24 x 103m apart and then released from rest. The initial acceleration of the first particle is obs

inessss [21]

Answer:

Part a)

$m_2 = 4.9 \times 10^7 kg$

Part b)

$q_1 = q_2 = 5312.6 C$

Explanation:

Part a)

As we know that both charge particles will exert equal and opposite force on each other

so here the force on both the charges will be equal in magnitude

so we will have

$F = m_1a_1 = m_2a_2$

here we have

$6.3 \times 10^7(7) = m_2(9)$

now we have

$m_2 = 4.9 \times 10^7 kg$

Part b)

Now for the force between two charges we can say

$F = \frac{kq_1q_2}{r^2}$

now we have

$(6.3 \times 10^7)(7) = \frac{(9\times 10^9)q^2}{(24\times 10^3)^2}$

now we have

$q_1 = q_2 = 5312.6 C$

3 0

3 years ago

A Imagine you derive the following expression by analyzing the physics of a particular system: v2 + 2az. The problem requires so

Nuetrik [128]

Answer:

z = 93.2 m

Explanation:

We can appreciate that this expression is equivalent to the linear motion equation with constant acceleration

v² = v₀² + 2 a d

If we make a term-to-term comparison with the expression obtained, they are equivalent

u² = v² + 2 a z

From here we can clear the position

2 a z = u² –v²

z = (u² –v²) / 2 a

Let's calculate

For the speed to reduce the acceleration must be negative

z = (0 - 21.8²) / 2(- 2.55)

z = 93.2 m

7 0

2 years ago