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Murljashka [212]
4 years ago
8

At one time, it was thought that positive charge was spread throughout the atom. This was the plum-pudding model. Rutherford's e

xperiment to test this theory involved shooting positively charged particles toward the atom. Describe what the results of this experiment would have been if the plum-pudding model were correct.
Physics
2 answers:
laiz [17]4 years ago
7 0
The 2 positively charged particles would repel each other.
dezoksy [38]4 years ago
4 0

Answer:

1) All the alpha particles must have deviated at some angle from their initial path.

2) None of the alpha particle will go out undeviating

Explanation:

In Rutherford's experiment he bombarded alpha particles on the nucleus of gold atom and then he observed that most of the alpha particles moves out undeviating while few of them shows small deflection.

But the most shocking part of experiment was that he saw that one out of 10,000 alpha particles are such which retrace their initial path i.e. those alpha particles are deviated nearly 180 degree angle.

Now these all observations will concluded as

1. Most of the part of atom is vacant and hence atom is not made up of positively charged.

2. Electrons revolves around the nucleus in circular orbits

3. The positive charge of the nucleus is concentrated at the nucleus of of atom which is a very small and dense volume of atom.

So these results are totally opposite of plum pudding model because as per plum pudding model the atom is a positively charged sphere and electrons are randomly placed in it.

If this model is assumed to be correct then in that case the observations must be like following

1) All the alpha particles must have deviated at some angle from their initial path.

2) None of the alpha particle will go out undeviating

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A river flows due south with a speed of 2.0 m/s. You steer a motorboat across the river; your velocity relative to the water is
neonofarm [45]

Answer:

a) 25.5°(south of east)

b) 119 s

c) 238 m

Explanation:

solution:

we have river speed v_{r}=2 m/s

velocity of motorboat relative to water is v_{m/r}=4.2 m/s

so speed will be:

a) v_{m}=v_{r}+v_{m/r}

solving graphically

v_{m}=\sqrt{v^2_{r}+v^2_{m/r}}

     =4.7 m/s

Ф=tan^{-1} (\frac{v_{r}}{v_{m/r}} )

  =25.5°(south of east)

b) time to cross the river: t=\frac{w}{v_{m/r}}=\frac{500}{4.2}=119 s

c) d=v_{r}t=(2)(119)=238 m

note :

pic is attached

6 0
4 years ago
A uniform metal bar 100cm long balances at 20cm mark when a mass of 1.5kg is attached at 0cm mark calculate the weight of the ba
beks73 [17]

Answer:

30 N

Explanation:

there are two forces act on the bar:

- weight of 1.5 kg mass, w = mg = 15 N

- weight of the bar, wb

for balance,

w * Lw = wb * Lwb

Lw = length of bar from the mass to the pivot

Lwb = lenght of bar from the center of the bar to the pivot

15 * 20 = wb * (50-20)

300 = wb * 30

wb = 300/30 = 30 N

4 0
4 years ago
If a hiker that weighs 600 newtons climbs a 50 meter hill, how much gravitational potential energy has the hiker gained?
Illusion [34]

The gravitational potential energy U is defined as the product of mass m, the acceleration of gravity g and the height of object h.

U = mgh

We do not have the mass of the hiker. But we know that its W weight is:

W = mg

Where

g = 9.8 m/s^2

So:

m = \frac{W}{g}\\\\m = \frac{600}{g}.

So:

U = (\frac{600}{g})gh\\\\U = (600N)(50m)

U = 30000J

The hiker has gained 30,000 J of energy

6 0
3 years ago
With what tension must a rope with length 2.90 m and mass 0.125 kg be stretched for transverse waves of frequency 42.0 Hz to hav
Vesnalui [34]

Answer:

41.64 N

Explanation:

Applying,

v = √(T/m')................ Equation 1

Where v = velocity of the wave, T = Tension of the rope, m' = mass per unit length of the rope.

make T the subject of the equation,

T = v²m'................. Equation 2

But,

v = λf............... Equation 3

Where λ = wavelength, f = frequency

And

m' = m/L........... Equation 4

Where m = mass of the rope, L = length of the rope

Substitute equation 3 and equation 4 into equation 2

T = (λf)²(m/L).............. Equation 5

From the question,

Given: λ = 0.740 m, f = 42 Hz, m = 0.125 kg, L = 2.9 m

Substitute these values into equation 5

T = (42×0.74)²(0.125/2.9)

T = 41.64 N

6 0
3 years ago
The metal cover that seals the top of the cylinder
padilas [110]

Answer:a lid

Explanation bob pulled the lid off the jar of pickles

5 0
3 years ago
Read 2 more answers
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