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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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Question # 40
jenyasd209 [6]

Answer:

Potential Energy = x = m g h

Kinetic energy = 1/2 m v^2

Assuming the mass fall from rest

1/2 m v^2 = m g h

v^2 = 2 g h

So the speed attained is independent of the mass

Also, x / v   does not have the units of mass

So the solution is none of the above.

7 0
3 years ago
An object becomes positively charged when it
Montano1993 [528]

Answer:

D

Explanation:

8 0
4 years ago
What is the magnitude of the resultant vector? Round
Naddika [18.5K]

Answer: 13.9 m

Explanation:

4 0
3 years ago
Josh starts throwing his pitch; we assume that his arm moves in a perfect circular arc, and his arm is 0.7 m long. If he needs t
harina [27]

As we know that centripetal acceleration is given as

a_c = \frac{v^2}{R}

here we know that

linear speed = 35 m/s

here radius of path is same as length of arm

R = 0.7 m

now from above above formula we have

a_c = \frac{35^2}{0.7}

a_c = 1750 m/s^2

so acceleration is 1750 m/s/s

5 0
4 years ago
Which of the following statements about electromagnetic waves in free space are true? (There could be more than one correct choi
Vanyuwa [196]

Answer:

The electric field carries the same mount of energy as the magnetic field

The frequency of the magnetic field is the same as the frequency of the electric field.

Explanation:

Let's analyze each option in detail:

The electric and magnetic fields have equal amplitudes. --> FALSE. In fact, the amplitudes of the electric and magnetic field are related by the equation:

E=cB

where

E is the amplitude of the electric field

B is the amplitude of the magnetic field

c is the speed of light

Therefore, from the equation we see that the amplitude of the electric field is much larger than that of the magnetic field.

The electric field carries the same mount of energy as the magnetic field. --> TRUE.

The energy carried by the electric field is:

u_E = \frac{1}{2}\epsilon_0 E^2

where \epsilon_0 is the vacuum permittivity.

The energy carried by the magnetic field is:

u_B = \frac{1}{2\mu_0}B^2

where \mu_0 is the vacuum permeability.

Given the following relationship:

c=\frac{1}{\sqrt{\epsilon_0 \mu_0}}

We can write

E=cB=\frac{1}{\sqrt{\epsilon_0 \mu_0}}B

So

u_E = \frac{1}{2}\epsilon_0 (\frac{1}{\sqrt{\epsilon_0 \mu_0}}B)^2=\frac{1}{2\mu_0}B^2

which means u_E=u_B.

The electric field carries more energy than the magnetic field. --> FALSE, because in disagreement with the calculations above.

The frequency of the magnetic field is the same as the frequency of the electric field. --> TRUE. In an electromagnetic waves, electric field and magnetic field oscillate at the same frequency.

The frequency of the electric field is higher than the frequency of the magnetic field. --> FALSE, because in disagreement with the previous statement.

8 0
3 years ago
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