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g100num [7]
3 years ago
5

The smallest separation resolvable by a microscope is of the order of magnitude of the wavelength used. What energy electrons wo

uld one need in an electron microscope to resolve separations of
(a) 150 Angstrom

(b) 5 Angstrom
Physics
1 answer:
hjlf3 years ago
6 0

Answer: a) for 150 Angstroms 6.63 *10^-3 eV; b) for 5 Angstroms 6.02 eV

Explanation: To solve this problem we have to use the relationship given by De Broglie as:

λ =p/h where p is the momentum and h the Planck constant

if we consider the energy given by acceleration tube for the electrons given by: E: e ΔV so is equal to kinetic energy of electrons p^2/2m

Finally we have:

eΔV=p^2/2m= h^2/(2*m*λ^2)

replacing we obtained the above values.

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The correct choice is

B. Particles at the bottom of the water carry heat energy to the top of the water.

when pot of water is heater, the bottom of pot gets heated. the particles of water in contact with the bottom of the pot gets heat through conduction. after getting heat, these particles of water near the bottom, move away towards top and their position is taken by cooler particles from top. that way heat travels

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3 years ago
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Suppose that you measure a galaxy's redshift, and from the redshift you determine that its recession velocity is 30,000 (3×10^4)
anyanavicka [17]

Answer:

1.4 billion light years away

Explanation:

v = Recessional velocity = 30000 km/s[/tex]

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According to Hubble's law

v=H_0D\\\Rightarrow D=\frac{v}{H_0}\\\Rightarrow D=\frac{3\times 10^{4}}{65}\times 3.2\times 10^6\\\Rightarrow D=1476923076.92307\ ly\\\Rightarrow D=1.4\times 10^9\ ly

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5 0
3 years ago
Electrical systems are governed by Ohm’s law, which states that V = IR, where V is the voltage, I is the current, and R is the r
ella [17]

Answer:

\frac{dR(t)}{dt}=0.06\Omega

Explanation:

Since R(t)=\frac{V}{I(t)}, we calculate the resistance rate by deriving this formula with respect to time:

\frac{dR(t)}{dt}=\frac{d}{dt}(\frac{V}{I(t)})=V\frac{d}{dt}(\frac{1}{I(t)})

Deriving what is left (remember that (\frac{1}{f(x)})'=-\frac{1}{f(x)^2}f'(x)):

\frac{d}{dt}(\frac{1}{I(t)})=-\frac{1}{I(t)^2}\frac{dI(t)}{dt}

So we have:

\frac{dR(t)}{dt}=-\frac{V}{I(t)^2}\frac{dI(t)}{dt}

Which for our values is (the rate of <em>I(t)</em> is decreasing so we put a negative sign):

\frac{dR(t)}{dt}=-\frac{24V}{(56A)^2}(-8A/s)=0.06\Omega

8 0
3 years ago
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