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3 years ago

Trình bày về Thí nghiệm tán xạ Rutherford, phát hiện proton.

Physics

1 answer:

Setler [38]3 years ago

6 0

Answer:

Any other language I don't know this language

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KatRina [158]

Answer:

Antarctica and Australia were one landmass millions of years ago

Explanation:

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5 0

3 years ago

Read 2 more answers

Cousin Throckmorton is playing with the clothesline. One end of the clothesline is attached to a vertical post. Throcky holds th

Oksi-84 [34.3K]

Answer:

The frequencies are $f_n = n (0.875 )$

Explanation:

From the question we are told that

The speed of the wave is $v = 0.700 \ m/s$

The length of vibrating clothesline is $L = 40.0 \ cm = 0.4 \ m$

Generally the fundamental frequency is mathematically represented as

$f = \frac{v}{2 L }$

=> $f = \frac{ 0.700 }{2 * 0.4 }$

=> $f = 0.875 \ Hz$

Now this other frequencies of vibration experience by the clotheslines are know as harmonics and they are obtained by integer multiple of the fundamental frequency

The frequencies are mathematically represented as

$f_n = n * f$

=> $f_n = n (0.875 )$

Where n = 1, 2, 3 ....

3 0

3 years ago

You observe a plane approaching overhead and assume that its speed is 600 miles per hour. The angle of elevation of the plane is

scZoUnD [109]

Answer:4.34 miles

Explanation:

first Elevation = $19^{\circ}$

After 1 minute Elevation changes to $59^{\circ}$

Ditsance travelled in 1 minute = $600\times \frac{1}{60}$ =10 mile

Now

tan59= $\frac{H}{x}$

H=xtan59

tan19= $\frac{H}{10+x}$

H= $\left ( 10+x\right )tan19$

Equating H

we get

1.319x=10tan19

x=2.61 miles

H= $2.61\times tan59$ =4.34 miles

4 0

3 years ago

What is the relation between the wavelength, λ, and the momentum, p, of a particle? Roughly what kinetic energy should an electr

prisoha [69]

Answer:

λ = h/p

$E=2.41*10^{-17}Joules$

This energy is equivalent to E=150.44eV

Explanation:

de Broglie Equation:

λ = h/p

h=6.63×10-34 m2 kg / s Planck's constant

so: p=h/λ

On the other hand, an atomic distance, an angstrom, is equal: $10^{-10}m$ , λ should be similar to this distance.

Kinetics Energy of a electron, m=9.11×10−31 kg

$E=\frac{p^{2}}{2m_e}=\frac{h^{2}}{2\lambda^{2}m_e} =\frac{(6.63*10^{-34})^{2}}{2*(10^{-10})^{2}*9.11*10^{-31}}=2.41*10^{-17}Joules$

Number of volts needed to accelerate that electron from rest to this energy:

$E=Vq_{e}$

$V=E/q_{e}=2.41*10^{-17}/(1.602*10^{-19})=150.44Volts$

So, this energy is equivalent to E=150.44eV

This high-voltage device is called: particles accelerator

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3 years ago

Our school needs to offer healthier options in the lunchroom. Elever High School has recently updated its cafeteria menu to incl

White raven [17]

Answer:

Explanation:

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3 years ago