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Phantasy [73]
3 years ago
10

J.J Thompson proposed that atoms were made up of positively charged particles with electrons scattered into them. How did he rea

ch this conclusion?
Physics
1 answer:
Harlamova29_29 [7]3 years ago
8 0
First, J.J Thompson experimented with cathode ray tubes: sealed glass tubes at vaccum (without air inside), with two electrons (cathode and anode). When a high voltage was applied a beam of particles left from the cathode and passed throuhg two charged plates (one negative and one positive).

The beam of particles was deflected toward the positively-charged electric plate, which indeicatedd that the particles were negatively charged.

Given that Thompson experimented with different materials (cathode electrodes), that the results were always the same, and that the mass of the electrons were a small fraction (approximately 1/2000) of the hydrogen atom, he concluded that the negatively particle was part of the atom (a subatomic particle).

Also, given that the atom is neutral, he concluded that there were negative and positive particles in any atom, and he speculated that the negative particles (electrons) were scattered into the positive particles which were way more massive.


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What is mass?Explain
Alexxx [7]

Answer:

The term mass is used to refer to the amount of matter in any given object For instance, a person or object may be weightless on the moon because of the lack of gravity but that same person or object maintains the same mass regardless of location

Explanation:

8 0
3 years ago
What is the answer for ×^+2×+ with solustion​
drek231 [11]
There’s something wrong with this equation!!! Verify the equation!!!
4 0
2 years ago
A certain wire has a resistance of 110 Ω. What is the resistance of a second wire, made of the same material, that is 1/4 as lon
Svet_ta [14]

Answer:

New Resistance = 247.5 ohm

Explanation:

Resistance = resistivity * length / area

Since resistivity for the material is constant, resistance is directly proportional to (length/area).

This means that if (length/area) decreases or increases by any ratio, then resistance will increase or decrease by the same ratio.

So let's find the change in length/area

New length = 0.25 old length

New area = (1/9) old area                                 (This is because area equation contains a square of the diameter. if diameter decreases by 1/3, area decreases by (1/3)^2   )

So we now get length /area:

New length / new area = ( 0.25 old length) / (1/9 of old area)

New length / new area = 9*0.25 (old length / old area)

New length / new area = 2.25 (old length / old area)

To get the new resistance, we simply multiply it by the ratio we just found.

This equals:

110 * 2.25 = 247.5 ohm

4 0
3 years ago
Read 2 more answers
Laura adds 50mL of boiling water to 100mL of ice water. If the 150 mL of water is then put into a freezer, at what temperature w
Dennis_Churaev [7]
Water that is pure, will always freeze at 0 degrees.
8 0
2 years ago
A pump is used to send water through a hose, the diameter of which is 10 times that of the nozzle through which the water exits.
Keith_Richards [23]

Use VFR1 = VFR2 to discover the velocity at in the hose VFR = A * V

D hose =10 * D nozzle, R hose = 5 * D nozzle

Area of a circle = πR^2

Area h=3.14*25*D^2 = 75.5D^2

(Radius=Diameter/2) area n = 3.14*(D^2/4) = .785D^2

 

Use VFR = VFR v2 = 0.4m/s

0.4*.785D^2 = 75.5*D^2* v1 D^2

= .314 =75.5*V1

v1 = 0.004m/s

 

Now we have the velocity, we can use Bernoulli's equation.

P1+ρgh1+ρV1^2 /2 = constant

There is no atmospheric pressure before so the P1= the gauge pressure at the pump, let’s call the height of the hose 0m and the height of the nozzle 1m so the is no ρgh1 Likewise, there is only atmospheric pressure at the nozzle which is 100000 PA, and lastly the density ρ of water is 1000 KG/M^3

Pg + 1000*.004^2/2 = 100000+1000*9.8*1+ 1000*0.4^2/2

Pg + .008= 100000+9800+80

Pg+.008= 109880

Pg=109880.008 PA

3 0
3 years ago
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