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Salsk061 [2.6K]

3 years ago

14

Which activity best demonstrates Ernest Rutherford's creativity?A. questioning the correctness of the plum pudding modelB. desig

ning an experiment to test the plum pudding modelC. making observations about the gold foil experimentD. summarizing the conclusions from the gold foil experiment as a theory

1 answer:

Rus_ich [418]3 years ago

5 0

Answer:

C.

Explanation:

Of the following activity best demonstrates the creativity of Rutherford: is making observations about the gold foil experiment.

Rutherford's Gold Foil Test confirmed the existence of a thin, enormous atom core that would later be recognized as an atom's nucleus. To examine the effect of alpha particles on material, Ernest Rutherford, Hans Geiger and Ernest Marsden conducted their Gold Foil Experiment.

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Aluminum allows for the flow of electrons, while glass does not. Would an aluminum wire surrounded by glass be an effective desi

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answer

no

Explanation:

I do not think that I would because even though its a conductor in the insulator I think it would insulate it before it will work (not sure if that makes sense)

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2 years ago

An electron is confined in a harmonic oscillator potential well. What is the longest wavelength of light that the electron can a

kipiarov [429]

Answer:

The longest wavelength of light is 209 nm.

Explanation:

Given that,

Spring constant = 74 N/m

Mass of electron $m= 9.11\times10^{-31}\ kg$

Speed of light $c= 3\times10^{8}\ m/s$

We need to calculate the frequency

Using formula of frequency

$f =\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

Where, k= spring constant

m = mass of the particle

Put the value into the formula

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{74}{9.11\times10^{-31}}}$

$f=1.434\times10^{15}\ Hz$

We need to calculate the longest wavelength that the electron can absorb

$\lambda=\dfrac{c}{f}$

Where, c = speed of light

f = frequency

Put the value into the formula

$\lambda =\dfrac{3\times10^{8}}{1.434\times10^{15}}$

$\lambda=2.092\times10^{-7}\ m$

$\lambda=209\ nm$

Hence, The longest wavelength of light is 209 nm.

6 0

3 years ago

Water at 25°C and 1 atm is flowing over a long flat plate with a velocity of 14 m/s. Determine the distance from the leading edg

Gre4nikov [31]

Answer:

L=31.9 mm

δ = 0.22 mm

Explanation:

Given that

v= 14 m/s

ρ=997 kg/m³

μ= 0.891 × 10⁻3 kg/m·s

As we know that when Reynolds number grater than 5 x 10⁵ then flow will become turbulent.

$Re=\dfrac{\rho vL}{\mu}$

$L=\dfrac{Re\mu}{\rho v}$

$L=\dfrac{5\times 10^5\times 0.891\times 10^{-3}}{ 14 \times 997}\ m$

L=0.0319 m

L=31.9 mm

The thickness of the boundary layer at that location L given as

$\delta =\dfrac{5L}{\sqrt{Re}}$

$\delta =\dfrac{5\times0.0319}{\sqrt{5\times 10^5}}$

δ = 0.00022 m

δ = 0.22 mm

7 0

2 years ago

Consider a concave mirror that has a focal length f. In terms of f, determine the object distances that will produce a magnifica

Aleks04 [339]

We have that the magnification of each focal length is given respectively as

A) has $u=3\frac{f}{2}$

B) has $u=4\frac{f}{3}$

C) has $u=5\frac{f}{4}$

From the question we are told that:

Focal Length F

Generally, the equation for Magnification is mathematically given by

$M=\frac{-v}{u}$

Therefore

$v=2u$

For A

$M=-2$

Therefore

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$\frac{1}{f}=\frac{1}{u}+\frac{1}{2u}$

Therefore

$u=3\frac{f}{2}$

For B

$M=-3$

Therefore

$v=3u$

Where

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$\frac{1}{f}=\frac{1}{u}+\frac{1}{3u}$

Therefore

$u=4\frac{f}{3}$

For C

$M=-4$

Therefore

$v=4u$

Therefore

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

$\frac{1}{f}=\frac{1}{u}+\frac{1}{4u}$

Therefore

$u=5\frac{f}{4}$

Conclusion

From the calculations above we can rightly say that the magnifications of the values above are

A has $u=3\frac{f}{2}$

B has $u=4\frac{f}{3}$

C has $u=5\frac{f}{4}$

For more information on this visit

brainly.com/question/14468351

3 0

2 years ago