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MrRa [10]
3 years ago
7

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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Which mixtures could be separated using evaporation? Check all that apply. fat in milk salt in ocean water mud in pond water but
salantis [7]
I would go with salt in ocean water as when you heat it and the water begins to evaporate it will leave the salt behind


Also water in mud as the water would evaporate and leave the mud residue behind


GOOD LUCK
BRAINLIEST IF HELPED
5 0
3 years ago
Read 2 more answers
Si pudieras viajar a la Luna:
Nostrana [21]

Answer:

i) Distancia, ii) La cinta métrica es impracticable.

Explanation:

i) El concepto físico que se construye únicamente del punto de salida y el punto de llegada a la Luna es el concepto de desplazamiento, definido como la distancia en línea recta de un punto en el espacio con respecto a un punto de referencia (la Tierra en este caso).

La distancia puede involucrar trayectorias curvilíneas entre los puntos mencionados.

ii) Por último, el uso de una cinta métrica es impracticable debido a la cantidad de material a utilizar y los efectos gravitacionales, electromagnéticos y mecánicos que inducen a una deflexión o una ruptura de esa cinta debido a la magnitud de la distancia entre las superficies del planeta y el satélite, respectivamente.

En este caso, es mejor utilizar la medición con tecnología láser, basadas en el fenómeno del electromagnetismo.

4 0
3 years ago
Can someone give me an example or explanation of calculating energy from voltage? Many thanks!
antoniya [11.8K]
Expression to calculate energy from voltage: E= V*Q where E= energy, V= voltage, and Q= charge

Additional help:
-To find the Voltage ( V )
[ V = I x R ] V (volts) = I (amps) x R (Ω)

-To find the Current ( I )
[ I = V ÷ R ] I (amps) = V (volts) ÷ R (Ω)

-To find the Resistance ( R )
[ R = V ÷ I ] R (Ω) = V (volts) ÷ I (amps)

I hope that helps to some extent-
7 0
3 years ago
A 0.l ‑kilogram block is attached to an initially unstretched spring of force constant k = 40 N/m as shown right. The block is d
GalinKa [24]

Answer:

The maximum potential energy of the system is 0.2 J

Explanation:

Hi there!

When the spring is stretched, it acquires potential energy. When released, the potential energy is converted into kinetic energy. If there is no friction nor any dissipative forces, all the potential energy will be converted into kinetic energy according to the energy conservation theorem.

The equation of elastic potential energy (EPE) is the following:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = stretching distance.

The elastic potential energy is maximum when the block has no kinetic energy, just before releasing it.

Then:

EPE = 1/2 · 40 N/m · (0.1 m)²

EPE = 0.2 J

The maximum potential energy of the system is 0.2 J

8 0
3 years ago
A convex spherical mirror having a radius of curvature 18 cm (focal length = 1/2 radius of curvature for a spherical mirror) pro
tekilochka [14]

Answer:

distance between object and image =  18.9 cm

Explanation:

given data

radius of curvature = 18 cm

focal length = 1/2 radius of curvature

magnification = 40%

to find out

distance between object and image

solution

we know lens formula that is

1/f = 1/v + 1/u     ....................1

here f = 18 /2 and v and u is object and image distance

and we know m = 40% = 0.40

so 0.40 = -v / u

so here v = - 0.40 u

so from equation 1

1/f = 1/v + 1/u

2/18 = - 1/0.40u + 1/u

u = -13.5 cm   ..................2

and

v = -0.40 (- 13.5)

v = 5.4 cm     ......................3

so from equation 2 and 3

distance between object and image =  5.4 + 13.5

distance between object and image =  18.9 cm

6 0
3 years ago
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