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3 years ago

Inside a freely falling runaway elevator, your

Physics

2 answers:

nasty-shy [4]3 years ago

7 0

The answer is apparent weight is zero.

You are still accelerating downwards at 9.8m/s^2 (if you are on Earth).

You still are being affected by the Earth's gravity.

Not all because of the previous two statements.

Not none because apparent weight is zero as you are falling.

umka21 [38]3 years ago

5 0

Inside a freely falling runaway elevator, your <em>apparent weight is zero,</em> and you have no way to tell that you're inside a freely falling elevator.

Your ACTUAL acceleration is the acceleration of gravity (9.8 m/s² down), but you have no way to test it or measure it.

All you can tell is that you feel weightless. Any experiment you can do inside the elevator shows you that Newton's laws of motion and gravity are all true, and that electromagnetic waves behave normally.

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A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it aroun

nalin [4]

Answer:

Therefore the the highest frequency is 620Hz and lowest frequency is 580Hz

Explanation:

Given data

Source Frequency fs=600Hz

Length r=1.0m

RPM=100 rpm

The speed of the generator is calculated as:

$v_{s}=rw\\v_{s}=r(2\pi f)$

Substitute the given values

$v_{s}=(1.0m)2\pi (\frac{100}{60}rev/s )\\v_{s}=10.47m/s$

For approaching generator the frequency is calculated as:

$f_{+}=\frac{f_{s}}{1-\frac{v_{s}}{v} }\\f_{+}=\frac{600Hz}{1-\frac{10.47m/s}{343m/s} } \\f_{+}=620Hz$

On the other hand,for the receding generator,Doppler's effect is expressed as:

$f_{-}=\frac{f_{s}}{1+\frac{v_{s}}{v} }\\f_{-}=\frac{600Hz}{1+\frac{10.47m/s}{343m/s} } \\f_{-}=580Hz$

Therefore the the highest frequency is 620Hz and lowest frequency is 580Hz

8 0

3 years ago

Doubly ionized lithium Li2+ (Z = 3) and triply ionized beryllium Be3+ (Z = 4) each emit a line spectrum. For a certain series of

EleoNora [17]

Answer:

tex]\lambda_{Be}[/tex] = 22.78 nm

Explanation:

Bohr's model for the hydrogen atom has been used by other atoms with a single electric charge by changing the number of charges by the charge of the new atom (atomic number)

$E_{n}$ = k e² / 2a₀ (1 /n²)

ao = h'² / k m e² h' = h/2πi

For another atom with a single electron in the last layer

a₀ ’= h’² / k m (Ze)²

a₀ ’= a₀ / Z²

Therefore, when replacing in the equation

$E_{n}$ = - Z² Eo/n²

E₀ = 13,606 eV

The transition occurs when the electron stops from one level to another

$E_{n}$ - $E_{m}$ = Z² E₀ (1 / n² - 1 / m²) = Z² ΔE

Let's relate this expression to the wavelength

c = λ f

E = h f

E = h c /λ

h c / λ = Z² ΔE

λ = 1 / Z² (hc / ΔE)

λ = 1 / Z² λ_hydrogen

Let's apply this last equation to our case

Lithium Z = 3

$E_{n}$ = - 9 Eo / n²

40.5 10-9 = 1/9 λ_hydrogen

Beryllium Z = 4

λ = 1/16 λ_hydrogen

Let's write our two equations is and solve

40.5 10-9 = 1/9 λ_hydrogen

tex]\lambda_{Be}[/tex] = 1/ 16 λ_hydrogen

40.5 10⁻⁹ = 1/9 (16 $\lambda_{Be}$ )

tex]\lambda_{Be}[/tex] = 40.5 9/16

tex]\lambda_{Be}[/tex] = 22.78 nm

6 0

3 years ago

Which image illustrates the bouncing of a light wave off of a surface?

Naily [24]

Answer:

The answer is A good luck :P

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Hat is the approximate pressure of air at sea level?

Tanya [424]

c) 101kPa

Hope I helped! ( Smiles )

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3 years ago

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The atmosphere protects us from _____ and _____.

sveticcg [70]

Answer: B. meteorites, UV rays

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7 0

3 years ago

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