( 45 points)

2 answers:

6 0

The British physicist Joseph John (J. J.) Thomson (1856–1940) performed a series of experiments in 1897 designed to study the nature of electric discharge in a high-vacuum cathode-ray tube, an area being investigated by many scientists at the time. Thomson's model showed the atom as a positively charged ball of matter with negatively changed electrons floating freely around inside of it. This model showed the atom having no structure. There are also no protons and neutrons in this model. Thomson knew that the atom had positively and negatively charges particles in it he just didn't know how they were arranged. Today's model gives us a much clearer picture of the atom. There is a positively charged center of the atom that is denser than the rest of it called the nucelus. This dense center is made up of positively charged protons and neutrally charged neutrons. Around the outside of the nucleus the electrons are organized on rings. These electrons are arranged in a certain pattern that is the same for all atoms.

alekssr [168]3 years ago

5 0

J.J. Thomson's experiments with cathode ray tubes showed that all atoms contain tiny negatively charged subatomic particles or electrons.Thomson's plum pudding model of the atom had negatively-charged electrons embedded within a positively-charged soup.Rutherford's gold foil experiment showed that the atom is mostly empty space with a tiny, dense, positively-charged nucleus.Based on these results, Rutherford proposed the nuclear model of the atom.

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A 0.200-kg cube of ice (frozen water) is floating in glycerine.The gylcerine is in a tall cylinder that has inside radius 3.90 c

natita [175]

Answer:

Part a)

h = 0.86 cm

Part b)

Level will increase

Explanation:

Part a)

Mass of the ice cube is 0.200 kg

Now from the buoyancy force formula we know that weight of the ice is counter balanced by buoyancy force on the ice

So here we will have

$mg = \rho V_{displaced} g$

$V_{displaced} = \frac{m}{\rho}$

$V_{displaced} = \frac{0.200}{1260} = 1.59 \times 10^{-4} m^3$

now as we know that ice will melt into water

so here volume of water that will convert due to melting of ice is given as

$V\rho_w = m_{ice}$

$V = \frac{0.200}{1000} = 2\times 10^{-4} m^3$

So here extra volume that rise in the level will be given as

$\Dleta V = V - V_{displaced}$

$\pi r^2 h = 2\times 10^{-4} - 1.59 \times 10^{-4}$

$(\pi (0.039^2) h = 0.41 \times 10^{-4}$

$h = 0.86 cm$

Part b)

Since volume of water that formed here is more than the volume that is displaced by the ice so we can say that level of liquid in the cylinder will increase due to melting of ice

5 0

3 years ago

Un hamster esta sentado sobre un tocadisco cuya rapidez angular es constante si el hamster se mueve a un punto localizado al dob

kotegsom [21]

Answer:

b) se duplica

Explanation:

The disk is moving with constant angular velocity, let's call it $\omega$ .