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Wewaii [24]
3 years ago
7

Oppositely charged objects attract each other. This attraction holds electrons in atoms and holds atoms to one another in many c

ompounds. However, Ernest Rutherford’s model of the atom failed to explain why electrons were not pulled into the atomic nucleus by this attraction. What change to the atomic model helped solve the problem seen in Rutherford’s model?
Physics
2 answers:
GalinKa [24]3 years ago
6 0

Answer and Explanation:

This can be explained as in Rutherford's model of atom the electrons orbits the nucleus which means that they will travel around the nucleus with some velocity and hence radiate electromagnetic waves which results in the loss of energy due to which the electron keeps coming closer and eventually falls into the nucleus.

But Bohr came up with a better explanation as according to the Bohr's atomic model, electrons stay fixed in orbit with certain energy in different shells around the nucleus and can only jump from an energy level to another if that specific amount of energy is supplied to it.

This model is based on the quantization of energy thus giving an explanation why electrons do not fall into the nucleus of an atom.

strojnjashka [21]3 years ago
6 0

Answer:

A) Bohr’s work with atomic spectra led him to say that the electrons were limited to existing in certain energy levels, like standing on the rungs of a ladder.

Explanation:

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Steel is an alloy of _____. One of the main reasons steel is used more widely than iron is because it’s_____ than iron. thanks f
yarga [219]

1. Iron I think and 2. stronger

8 0
3 years ago
How is an ammeter connected in a circuit to measure current flowing through it?
trasher [3.6K]

Answer:

It is connected in series with the circuit

Explanation:

This is because to measure the current in the circuit, the current in the circuit has to flow through the ammeter. As such, the ammeter must be connected in series with the circuit so as to measure the current flowing through the circuit.

So, to measure the current flowing through a circuit with an ammeter, the ammeter must be connected in series with the circuit.

4 0
3 years ago
Young's Modulus refers to changes in the a Volume b- Length c- Body layers
Novosadov [1.4K]

Explanation:

Young' modulus is the ratio of normal stress to the longitudinal strain. Mathematically, it is given by :

Normal stress is given by force per unit area. Longitudinal strain is the change in length per unit original length.

The mathematical definition of Young's modulus is given by :

Y=\dfrac{F/A}{\Delta L/L}..........(1)

Where

\Delta L is the change in length

F is the force

A is the area of cross section

So, the Young's modulus refers to the change in length of the object. Hence, the correct option is (b) "length".

8 0
3 years ago
2. Below what depth would a submarine have to submerge so that it would not be swayed by surface waves with a wavelength of 24 m
Travka [436]

<u>Answer:</u> Below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves

<u>Explanation:</u>

To avoid the surface waves, a submarine has to submerge below the wave base. It is the position below which the motion of the waves is negligible.

This wave base is equal to half of the wavelength. The equation becomes:

Wave base = \frac{\text{Wavelength}}{2}

We are given:

Wavelength = 24 m

Putting values in above equation, we get:

Wave base = \frac{24m}{2}=12m

Hence, below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves

4 0
3 years ago
A spring stretches 0.018 m when a 2.8-kg object is suspendedfrom its end. How much mass should be attached to this spring sothat
agasfer [191]

Answer:

m = 4.29 kg

Explanation:

Given that,

Mass of the object, m = 2.8 kg

Stretching in the spring, x = 0.018 m

Frequency of vibration, f = 3 Hz

Let m is the mass of the object that is attached to the spring. When it is attached the gravitational force is balanced by the force on spring. It is given by :

mg=kx

k=\dfrac{mg}{x}

k=\dfrac{2.8\times 9.8}{0.018}

k = 1524.44 N/m

Since, \omega=\sqrt{\dfrac{k}{m}}

m=\dfrac{k}{4\pi^2 f^2}

m=\dfrac{1524.44}{4\pi^2 \times 3^2}

m = 4.29 kg

So, the mass that is attached to this spring is 4.29 kg. Hence, this is the required solution.

4 0
3 years ago
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