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3 years ago

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What tool could help a biologist study the movements of cells? digital camera with a zoom lens a satellite with a digital camera

a television with a remote control a microscope with a video camera

2 answers:

Annette [7]3 years ago

4 0

I think your answer would be (D) microscope with a video camera
Hope i helped!

Mandarinka [93]3 years ago

4 0

Answer:

a microscope with a video camera

Explanation:

gg Dva style

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Consider a hydrogen atom in the n = 1 state. The atom is placed in a uniform B field of magnitude 2.5 T. Calculate the energy di

dlinn [17]

Answer:

$E=29\times 10^{-5}eV$

Explanation:

For n-=1 state hydrogen energy level is split into three componets in the presence of external magnetic field. The energies are,

$E^{+}=E+\mu B$ ,

$E^{-}=E-\mu B$ ,

$E^{0}=E$

Here, E is the energy in the absence of electric field.

And

$E^{+} and E^{-}$ are the highest and the lowest energies.

The difference of these energies

$\Delta E=2\mu B$

$\mu=9.3\times 10^{-24}J/T$ is known as Bohr's magneton.

B=2.5 T,

Therefore,

$\Delta E=2(9.3\times 10^{-24}J/T)\times 2.5 T\\\Delta E=46.5\times 10^{-24}J$

Now,

$Delta E=46.5\times 10^{-24}J(\frac{1eV}{1.6\times 10^{-9}J } )\\Delta E=29.05\times 10^{-5}eV\\Delta E\simeq29\times 10^{-5}eV$

Therefore, the energy difference between highest and lowest energy levels in presence of magnetic field is $E=29\times 10^{-5}eV$

6 0

3 years ago

The apparent reversal of motion of planets against the backdrop of fixed stars is called

Free_Kalibri [48]

The planets move eastward against the background of fixed stars with the exception of Venus, Uranus and Pluto moving westward as seen in Earth's sky . This apparent retreating movement is called Retrograde motion. It is illusion created by Earth's movement going by outer planets in their respective orbits.

7 0

3 years ago

Read 2 more answers

Find the magnitude of the average force ⟨Fx⟩⟨Fx⟩ in the x direction that the particle exerts on the right-hand wall of the conta

hodyreva [135]

Answer:

See explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem.

The attached file has a detailed solution of the given problem.

4 0

3 years ago

seraphim [82]

Answer:

ᵒᵒᵒᵒᵏᵏ

ᵏᵏᵏ

Explanation:

ᵒᵒᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏ

ᵏᵒᵒᵒᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏᵏᵐᵐ

6 0

3 years ago

An observer stands 24.7 m behind a marksman practicing at a rifle range. The marksman fires the rifle horizontally, the speed of

GaryK [48]

Explanation:

The given data is as follows.

Velocity of bullet, $c_{p}$ = 814.8 m/s

Observer distance from marksman, d = 24.7 m

Let us assume that time necessary for report of rifle to reach the observer is t and will be calculated as follows.

t = $\frac{24.7}{343}$ (velocity in air = 343 m/s)

= 0.072 sec

Now, before the observer hears the report the distance traveled by the bullet is as follows.

$d_{b} = c_{b} \times t$

= $814.8 \times 0.072$

= 58.66

= 59 (approx)

Thus, we can conclude that each bullet will travel a distance of 59 m.

8 0

3 years ago