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Jlenok [28]
3 years ago
10

Compare the characteristics of 4d orbitals and 3d orbitals and compared the following sentences. Check all that apply

Physics
1 answer:
charle [14.2K]3 years ago
8 0

Answer :

<em>(b) 4d orbitals would be larger in size than 3d orbitals</em>

<em>(e) 4d orbitals would have more nodes than 3d orbitals</em>

Explanation :

As we move away from one orbital to another, the distance between nucleus and orbital increases. So, 4d orbitals would be far to the nucleus than 3d orbitals.

Hence, 4d orbitals would be larger in size than 3d orbitals.

Number of nodes is any orbital is n - 1 where, n is principal quantum number.

So, number of orbital in 4d is 3.

And number of orbital in 3d is 2.

So, options (b) and (e) are correct.

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2. Below what depth would a submarine have to submerge so that it would not be swayed by surface waves with a wavelength of 24 m
Travka [436]

<u>Answer:</u> Below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves

<u>Explanation:</u>

To avoid the surface waves, a submarine has to submerge below the wave base. It is the position below which the motion of the waves is negligible.

This wave base is equal to half of the wavelength. The equation becomes:

Wave base = \frac{\text{Wavelength}}{2}

We are given:

Wavelength = 24 m

Putting values in above equation, we get:

Wave base = \frac{24m}{2}=12m

Hence, below 12m of depth, the submarine has to submerge so that it would not be swayed by surface waves

4 0
2 years ago
If a 6.1 A resistive load is connected by 2-conductor stranded 2-AWG copper wire to a source voltage of 125.2 V that is 358 ft a
polet [3.4K]

Answer:

124.86 V

Explanation:

We have to first calculate the voltage drop across the copper wire. The copper wire has a length of 358 ft

1 ft = 0.3048 m

358 ft = 109.12 m

The diameter of 2 AWG copper wire (d) = 6.544 mm = 0.006544 m

The area of the wire = πd²/4 = (π × 6.544²)/4 = 33.6 mm²

Resistivity of wire (ρ) = 0.0171 Ω.mm²/m

The resistance of the wire = \frac{\rho A}{l}=\frac{0.0171*109.12 }{33.6} =0.056\ ohm

The voltage drop across wire = current * resistance = 6.1 A * 0.056 ohm = 0.34 V

The voltage at end = 125.2 - 0.34 = 124.86 V

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3 years ago
Why does the surfaces of the moon, mars and earth vary in appearance
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