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3 years ago

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A new school is being built in a field of wildflowers. A class wanted to develop a research project to predict the effects of th

e new school on wildflower species. When would be the best time for the class to conduct the study?

1 answer:

kotykmax [81]3 years ago

4 0

A new school is being built in a field of wildflowers. A class wanted to develop a research project to predict the effects of the new school on wildflower species. Given the background of the class, then probably, the best time for them to do the study is during right after the school is being built. Through that, they can get some data how the school affects the flowers. So, this helps the flowers in the future.

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A photon of wavelength 7.33 pm scatters at an angle of 157° from an initially stationary, unbound electron. What is the de Brogl

Ann [662]

Answer:

4.63 p.m.

Explanation:

The problem given here can be solved by the Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=157^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos157^\circ ) m\\\lambda^{'}-\lambda=2.42(1-cos157^\circ ) p.m.$

Therfore,

$\lambda^{'}-\lambda=4.64 p.m.$

Here, the photon's incident wavelength is $\lamda=7.33pm$

So,

$\lambda^{'}=7.33+4.64=11.97 p.m$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

here, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda$ and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therfore,

$\lambda_{e}=\frac{7.33\times 11.97}{\sqrt{7.33^{2}+11.97^{2}-2\times 7.33\times 11.97\times cos157^\circ }} p.m.\\\lambda_{e}=\frac{87.7401}{18.935} = 4.63 p.m.$

This is the de Broglie wavelength of the electron after scattering.

8 0

4 years ago

The ink drops have a mass m = 1.00×10^−11 kg each and leave the nozzle and travel horizontally toward the paper at velocity v =

luda_lava [24]

Answer:

9.98 × 10⁻⁹ C

Explanation:

mass, m = 1.00 × 10⁻¹¹ kg

Velocity, v = 23.0 m/s

Length of plates D₀ = 1.80 cm = 0.018 m

Magnitude of electric field, E = 8.20 × 10⁴ N/C

drop is to be deflected a distance d = 0.290 mm = 0.290 × 10⁻³ m

density of the ink drop = 1000 kg/m^3

Now,

Time = $\frac{\textup{Distance}}{\textup{Velocity}}$

or

Time = $\frac{\textup{0.016}}{\textup{23}}$

or

Time = 6.9 × 10⁻⁴ s

Now, force due to the electric field, F = q × E

where, q is the charge

Also, Force = Mass × acceleration

q × E = 1.00 × 10⁻¹¹ × a

or

a = $\frac{q\times8.20\times10^4}{1\times10^{-11}}$

Now from the Newton's equation of motion

$d=ut+\frac{1}{2}at^2$

where,

d is the distance

u is the initial speed

a is the acceleration

t is the time

or

$0.290\times10^{-3}=0\times(6.9\times10^{-4})+\frac{1}{2}\times(\frac{q\times8.20\times10^4}{1\times10^{-11}})\times(6.9\times10^{-4})^2$

or

q = 9.98 × 10⁻⁹ C

4 0

3 years ago

What number is between 2000 and 2500

Kitty [74]

If its just a number 2,300 is an answer

6 0

4 years ago

A slit of width 2.0 μm is used in a single slit experiment with light of wavelength 650 nm. If the intensity at the central maxi

Valentin [98]

Answer:

The intensity at 10° from the center is 3.06 × 10⁻⁴I₀

Explanation:

The intensity of light I = I₀(sinα/α)² where α = πasinθ/λ

I₀ = maximum intensity of light

a = slit width = 2.0 μm = 2.0 × 10⁻⁶ m

θ = angle at intensity point = 10°

λ = wavelength of light = 650 nm = 650 × 10⁻⁹ m

α = πasinθ/λ

= π(2.0 × 10⁻⁶ m)sin10°/650 × 10⁻⁹ m

= 1.0911/650 × 10³

= 0.001679 × 10³

= 1.679

Now, the intensity I is

I = I₀(sinα/α)²

= I₀(sin1.679/1.679)²

= I₀(0.0293/1.679)²

= 0.0175²I₀

= 0.0003063I₀

= 3.06 × 10⁻⁴I₀

So, the intensity at 10° from the center is 3.06 × 10⁻⁴I₀

5 0

3 years ago

Help needed Can a body be accelerating if it is moving in circle

belka [17]

<em>Answer:</em>

<em>When </em><em>a </em><em>body </em><em>is </em><em>moving </em><em>on </em><em>a </em><em>circle </em><em>it </em><em>is </em><em>accelerating </em><em>because </em><em>centripetal </em><em>acceleration</em><em> </em><em>is </em><em>always </em><em>acting </em><em>on </em><em>it </em><em>towards </em><em>the </em><em>center.</em>

<em>Please </em><em>see</em><em> the</em><em> attached</em><em> picture</em><em>.</em><em>.</em><em>.</em>

<em>From </em><em>the </em><em>above </em><em>diagram,</em><em>we </em><em>can </em><em>say </em><em>the </em><em>acceleration</em><em> </em><em>is </em><em>always </em><em>acting </em><em>on </em><em>the </em><em>body </em><em>when </em><em>it </em><em>moves </em><em>in </em><em>a </em><em>circle.</em>

<em>Hope </em><em>this </em><em>helps.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>

5 0

3 years ago

Read 2 more answers