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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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irga5000 [103]
The answer is 167 pounds.
7 0
3 years ago
A kind of variable that a researcher purposely changes in investigation is
Ugo [173]

Answer:

independent variable

Explanation:

3 0
3 years ago
Which statements best describe science? Check all that apply. A)Science uses beliefs and opinions to construct explanations. B)S
Maru [420]

Answer:

Science is supported by facts and processes.

Science involves observation and experimentation.

Science continually changes and is constantly updated.

6 0
3 years ago
Read 2 more answers
In my trigonometry class, we were assigned a problem on Angular and Linear Velocity.
Rzqust [24]

1) 0.0011 rad/s

2) 7667 m/s

Explanation:

1)

The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:

\omega=\frac{\theta}{t}

where

\theta is the angular displacement of the object

t is the time elapsed

\omega is the angular velocity

In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is

\theta=2\pi rad

And the time taken is

t=95 min \cdot 60 =5700 s

Therefore, the angular velocity of the telescope is

\omega=\frac{2\pi}{5700}=0.0011 rad/s

2)

For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation

v=\omega r

where

v is the linear velocity

\omega is the angular velocity

r is the radius of the circular orbit

In this problem:

\omega=0.0011 rad/s is the angular velocity of the Hubble telescope

The telescope is at an altitude of

h = 600 km

over the Earth's surface, which has a radius of

R = 6370 km

So the actual radius of the Hubble's orbit is

r=R+h=6370+600=6970 km = 6.97\cdot 10^6 m

Therefore, the linear velocity of the telescope is:

v=\omega r=(0.0011)(6.97\cdot 10^6)=7667 m/s

4 0
3 years ago
Can some one help me ;-;
enyata [817]

Answer: The first answer for the first problem, and the 2nd answer for the second problem

Explanation: For the first one, if it is absolute zero, the molecules would not move at all.

For the second one, the temperature of the sample will increase due to the movement.

6 0
3 years ago
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