All categories

3 years ago

What did Bohr's model of the atom include that Rutherford's model did not have?

Physics

2 answers:

Paladinen [302]3 years ago

5 0

Answer:

Its nucleus

Explanation:

did it on edge2020

lara31 [8.8K]3 years ago

4 0

Answer:

The correct option is energy levels

Explanation:

Rutherford's model of an atom suggests that an atom has a tiny positively charged central mass (now called the nucleus) which is surrounded by electrons (negatively charged) in a cloud-like manner.

Bohr's model went a bit further than the Rutherford's model in describing an atom by suggesting that the electrons which surrounds in the nucleus travel in fixed circular orbits. This description by Bohr was able to describe the energy levels of orbitals which assumes that smallest orbitals have the lowest energy while the largest orbitals have the highest energy.

You might be interested in

45 joules were expended to move a box weighing 30 newtons. How many meters was it moved?

slava [35]

Answer:

1.5 meter is the answer

7 0

3 years ago

Calculate the amount of energy produced in joules by 100- watt light bulb lit for 2.5 hours.

9966 [12]

100/2.5 is 40.
40 is the energy that is being produced

4 0

4 years ago

Describe how air pressure is exerted on objects.

Delicious77 [7]

The same amount of pressure is exerted on all sides of the object equally. If you had 13 psi, then that means on the top of the object, 13 psi would be pushing in on the object. From the bottom, 13 psi would be pushing up on the object. Same goes for all the sides as well. The pressure pushes 'in' on the object.

5 0

3 years ago

Speed of light in a substance with refractive index of 1.70

makkiz [27]

Whats the question? ....

7 0

3 years ago

LOTS OF POINTSA rocket of mass 40 000 kg takes off and flies to a height of 2.5 km as its engines produce 500 000 N of thrust.

Svetradugi [14.3K]

Answer:

i) E = 269 [MJ] ii)v = 116 [m/s]

Explanation:

This is a problem that encompasses the work and principle of energy conservation.

In this way, we establish the equation for the principle of conservation and energy.

$E_{k1}+W_{1-2}=E_{k2}\\where:\\E_{k1}= kinetic energy at moment 1\\W_{1-2}= work between moments 1 and 2.\\E_{k2}= kinetic energy at moment 2.$

$W_{1-2}= (F*d) - (m*g*h)\\W_{1-2}=(500000*2.5*10^3)-(40000*9.81*2.5*10^3)\\W_{1-2}= 269*10^6[J] or 269 [MJ]$

At that point the speed 1 is equal to zero, since the maximum height achieved was 2.5 [km]. So this calculated work corresponds to the energy of the rocket.

Er = 269*10^6[J]

ii ) With the energy calculated at the previous point, we can calculate the speed developed.

$E_{k2}=0.5*m*v^2\\269*10^6=0.5*40000*v^2\\v=\sqrt{\frac{269*10^6}{0.5*40000} }\\ v=116[m/s]$

8 0

3 years ago