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3 years ago

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

Physics

2 answers:

zysi [14]3 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]3 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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yaroslaw [1]

Answer:

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Explanation:

4 0

3 years ago

A straight 1.20 m long conductor has a 2.00 A current travelling toward the East. Earth's magnetic field in this location is 4.9

KATRIN_1 [288]

Answer:

Force's magnitude $=1.176\,10^{-4}\,N$

Direction: down (towards the center of the Earth)

Explanation:

Recall that the magnetic force on a conductor of length L carrying a current I in a magnetic field B is given by the equation: $F=I\,L\,B$ in the case the magnetic field B and the direction of the current are at 90 degrees from each other (which is our case). The direction of the force will be given by the "right hand rule" associated with the vector product that defines this force.

Since the current is moving East, and the magnetic field of the Earth goes from North to South, the resultant Force vector will be pointing towards the Earth (and perpendicular to the plane defined by the current's direction and the magnetic field B)

The magnitude of the force, is given by the formula above, and given that all quantities to be considered are is SI units, it will result in Newtons (N):

$F=I\,L\,B\\F=2\,*\,1.2\,*\,4.9\,10^{-5}\,N\\F=1.176\,10^{-4}\,N$

8 0

4 years ago

A 190 g glider on a horizontal, frictionless air track is attached to a fixed ideal spring with force constant 160 N/m. At the i

laiz [17]

(a) Let x be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work W done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to x is

W = - (1/2 kx ² - 1/2 k (0.0400 m)²)

(note that x > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)

By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to x, so

W = ∆K = 0 - 1/2 m (0.835 m/s)²

Solve for x :

- (1/2 (160 N/m) x ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²

==> x ≈ 0.0493 m ≈ 4.93 cm

(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is

W = - 1/2 k (0.0400 m)²

If v is the glider's maximum speed, then by the work-energy theorem,

W = ∆K = 1/2 m (0.835 m/s)² - 1/2 mv ²

Solve for v :

- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) v ²

==> v ≈ 1.43 m/s

√(k/m) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz

3 0

2 years ago

Suppose you could fit 100 dimes, end to end, between your card with the pinhole and your dime-sized sunball. how many suns could

Naddika [18.5K]

Answer: 100 suns