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EleoNora [17]
3 years ago
6

20 PTS BRAINLIEST If Thomson's model of the atom was correct, what would Rutherford have seen in his experiments?

Physics
2 answers:
12345 [234]3 years ago
8 0

Answer:

The correct answer is A

Explanation:

This is because in J.J. Thompson model, there would be no nucleus to reflect the alpha particles. This means there would be no reflection of the particles, making the answer A

Sati [7]3 years ago
5 0
The correct option is A.
J.J Thompson proposed the plum model of atomic structure. The model states that the atom is a sphere of positively charged matter in which electrons are embedded.
Rutherford designed an experiment to verify the truth of this model. He set up a beam of alpha particles which are to be passed a thin gold foil. If the JJ model of atomic structure is correct, all of the alpha particles suppose to pass through the foil, but some of the alpha particles bounced back while those that passed through the foil emerged at different angles.
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Rube goldberg machine steps
tankabanditka [31]

Answer:

Rube Goldberg Machine is "a comically involved, complicated invention, laboriously contrived to preform a simple operation." 2. What are the 6 Simple Machines? A. The 6 Simple Machines are: wedge, screw, lever, wheel and axel, inclined plane and pulley.

3 0
3 years ago
A ball rolls off a desk at a speed of 3 m/s and lands .40 seconds later. How far from the base of the desk does the ball land?
Salsk061 [2.6K]

Is the velocity constant? Is there any friction?

3 meters per second

then after 40 seconds it must 3*40 = 120 meters

120 meters or 0.12 km if you will

7 0
3 years ago
8.) If a car moving at 50km/h skids 15m with locked brakes, how far does the same car moving at 100km/h
pantera1 [17]

(8) A car starting with a speed <em>v</em> skids to a stop over a distance <em>d</em>, which means the brakes apply an acceleration <em>a</em> such that

0² - <em>v</em>² = 2 <em>a</em> <em>d</em> → <em>a</em> = - <em>v</em>² / (2<em>d</em>)

Then the car comes to rest over a distance of

<em>d</em> = - <em>v</em>² / (2<em>a</em>)

Doubling the starting speed gives

- (2<em>v</em>)² / (2<em>a</em>) = - 4<em>v</em>² / (2<em>a</em>) = 4<em>d</em>

so the distance traveled is quadrupled, and it would move a distance of 4 • 15 m = 60 m.

Alternatively, you can explicitly solve for the acceleration, then for the distance:

A car starting at 50 km/h ≈ 13.9 m/s skids to a stop in 15 m, so locked brakes apply an acceleration <em>a</em> such that

0² - (13.9 m/s)² = 2 <em>a</em> (15 m) → <em>a</em> ≈ -6.43 m/s²

So the same car starting at 100 km/h ≈ 27.8 m/s skids to stop over a distance <em>d</em> such that

0² - (27.8 m/s)² = 2 (-6.43 m/s²) <em>d</em> → <em>d</em> ≈ 60 m

(9) Pushing the lever down 1.2 m with a force of 50 N amounts to doing (1.2 m) (50 N) = 60 J of work. So the load on the other end receives 60 J of potential energy. If the acceleration due to gravity is taken to be approximately 10 m/s², then the load has a mass <em>m</em> such that

60 J = <em>m g h</em>

where <em>g</em> = 10 m/s² and <em>h</em> is the height it is lifted, 1.2 m. Solving for <em>m</em> gives

<em>m</em> = (60 J) / ((10 m/s²) (1.2 m)) = 5 kg

(10) Is this also multiple choice? I'm not completely sure, but something about the weight of the tractor seems excessive. It would help to see what the options might be.

4 0
3 years ago
A ball is thrown upwards at an unknown speed. in a time of
alexandr402 [8]

Answer:

a)  Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a)  We have equation of motion s=ut+\frac{1}{2} at^2, where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8m/s^2

 Substituting

    6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s

 Initial speed of the ball = 14.45 m/s

 b) We have equation of motion v^2=u^2+2as, where v is the final velocity

   s = 6 m, u = 14.45 m/s, a = -9.8m/s^2

    Substituting

        v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s

  So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion v^2=u^2+2as, where v is the final velocity

   u = 14.45 m/s, v =0 , a = -9.8m/s^2

   Substituting

     0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m

  Maximum height reached = 10.65 m

8 0
3 years ago
Based on what you learn about jovian moons by watching the videos or reading your textbook, what types of evidence for recent or
Valentin [98]

Answer:

b. craters, river valleys feeding into surface lakes of very cold liquids

Explanation:

  • Jovian moons are the four largest satellites like the moon of the Jupiter ie the Lo, Europa, Ganymede, Callisto and were first seen by the galileo. They are amiugly the largest moons with radii larger than the dwarf planet.  
  • Lo has more than 400 active volcanoes and dotted more than 100 mountains and has an extremely thin atmosphere made up of sulfur dioxide. The Europa has deep oceans of liquid water, and the layer of ice, and are characteristic of the tidal heating.  
  • <u>While the surface of Callisto is heavily cratered and has salty liquid water.</u>
7 0
4 years ago
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