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viva [34]
4 years ago
5

Which term describes the minimum distance between two objects required to distinguish them as two separate object?

Physics
1 answer:
mihalych1998 [28]4 years ago
3 0
The answer is abbe’s resolution limit in which RL is equals to 0.612 the wavelength divided by NA. This is the minimum distance between objects when you can still see them as separate objects. In addition, the shorter the wavelength is better and the larger numerical aperture is better. <span />
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To see if your results are reasonable, you can compare the final velocity of the stone as it falls down unwinding the wire from
pentagon [3]

Answer:

The velocity of the stone is 2.57 m/s.

Explanation:

Given that

Height = 0.337 m

We need to calculate the velocity of the stone

Using equation of motion

v^2-u^2=2gh

Where, v = velocity of stone

u = initial velocity

g = acceleration due to gravity

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Put the value into the formula

v^2-0=2\times9.8\times 0.337

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7 0
4 years ago
A potential difference of 3.00 nV is set up across a 2.00 cm length of copper wire that has a radius of 2.00 mm. How much charge
Anvisha [2.4K]

The number of charge drifts are 3.35 X 10⁻⁷C

<u>Explanation:</u>

Given:

Potential difference, V = 3 nV = 3 X 10⁻⁹m

Length of wire, L = 2 cm = 0.02 m

Radius of the wire, r = 2 mm = 2 X 10⁻³m

Cross section, 3 ms

charge drifts, q = ?

We know,

the charge drifts through the copper wire is given by

q = iΔt

where Δt = 3 X 10⁻³s

and i = \frac{V}{R}

where R is the resistance

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ρ is the resistivity of the copper wire = 1.69 X 10⁻⁸Ωm

So, i = \frac{\pi(r)^{2}V  }{pL}

q = \frac{\pi(r^{2} )Vt }{pL}

Substituting the values,

q = 3.14 X (0.02)² X 3 X 10⁻⁹ X 3 X 10⁻³ / 1.69 X 10⁻⁸ X 0.02

q = 3.35 X 10⁻⁷C

Therefore, the number of charge drifts are 3.35 X 10⁻⁷C

3 0
3 years ago
A train can travel at 200 mph. How long will it take you to get to Chicago if it is 1800 miles away?
labwork [276]
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5 0
3 years ago
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In the final situation below, the 8.0 kg box has been launched with a speed of 10.0 m/s across a frictionless surface. Find the
Murljashka [212]

Answer:

the energy of the spring at the start is 400 J.

Explanation:

Given;

mass of the box, m = 8.0 kg

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Apply the principle of conservation of energy to determine the energy of the spring at the start;

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¹/₂ x 8 x 10² = Ux

400 J = Ux

Therefore, the energy of the spring at the start is 400 J.

8 0
3 years ago
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