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timurjin [86]
2 years ago
9

Most exceptions to the trend of decreasing radius moving to the right within a period occur in the __________.

Physics
1 answer:
Gelneren [198K]2 years ago
8 0

Answer:

Most exceptions to the trend of decreasing radius moving to the right within a period occur in the d-block.

Explanation:

  • In a period as we advance from left to right, the number of valence electrons in the same shell increases due to which the effective nuclear charge increases and thus the atomic size decreases.
  • In d-block atomic radius initially decreases then remains constant and increases towards the end.
  • As one moves from Sc (scandium) to Zn (zinc), the effective nuclear charge increases by a factor of 1, this is because:
  1. The number of electrons are low in the inner shell.
  2. The shielding power of d-orbital is low.
  3. Inter electronic repulsions will be operating at a value less than the nuclear charge, which will result in decrease in atomic radii.
  • As the number of electrons in the inner orbital increases the outer electrons repel one another which enables them to push away.
  • Although d-orbital has less shielding power, the number of electrons present in it are high. Hence, the electron-electron repulsive force becomes dominant, this results in an increase in the atomic radii.

Therefore, most exceptions to the trend of decreasing radius moving to the right within a period occur in the d-block.

Learn more about the periodic table here:

<u>brainly.com/question/9238898</u>

#SPJ4

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A sound has 13 crests and 15 troughs in 3 seconds. When the second crest is produced the first is 2cm away from the source? Calc
Yuki888 [10]

Answer:

The wavelength will be 4 cm, frequency will be 4.66 Hz and wave speed is 18.6 cm/sec

Explanation:

Given:

No. of crest = 13

No. of trough = 15

Time = 3 seconds

Hence, 1 crest or 1 trough = \frac{1}{2}\lambda

therefore,

13 C + 15 T = 28(\frac{1}{2}\lambda)

=14\lambda

Time given 3 seconds

  = \frac{3}{14}s

\nu= \frac{14}{3}

\nu= 4.6 Hz \approx 5 Hz

2 cm distance is travelled is time period

\lambda = 4 cm

Again wave will travel in 1 T = 4 cm

wave speed v =\lambda\times\nu

= 4\times\frac{14}{3}

= 18.6 cm/s

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A steel ball of mass 0.1 kg falls from a height of 1.8metres
murzikaleks [220]
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7 0
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Your starship, the Aimless Wanderer,lands on the mysterious planet Mongo. As chief scientist-engineer,you make the following mea
Setler [38]

Answer:

a)  M = 4,997 10²⁰ kg ,  b)   T = 1.43 10³ s

Explanation:

a) This exercise should be solved in several parts, let's start by calculating the acceleration of gravity of this planet from kinematics

          v = v₀ - a t

As it indicates that there is no atmosphere, the friction force is zero and the initial and final velocity have the same module, but the opposite direction

         a = (v₀ - v) / t

         a = (15 - (-15)) /9.00 = 30/9

         a = 3.33 m / s²

Now we use Newton's second law where force is the force of universal attraction

          F = m a

         G m M / r² = m a

         M = a r² / G

Let's calculate

         M = 3.33 (1.00 10⁵)² / 6.67 10⁻¹¹

         M = 4,997 10²⁰ kg

b) The period of the ship's orbit

In this case we have a centripetal acceleration

The radius of the orbit is the radius of the plant plus the height of the ship from the surface

         R = R_{m} + h

         R = 1 10⁵ + 2.00 10⁴

         R = 12 10⁴ m

         F = m a

        G m M / R² = m a

Centripetal acceleration is

         a = v² / R

The orbit is circular therefore the velocity module is constant, so we can use the equation of uniform motion, where the distance is the length of the orbit, for a circle

        d = 2π R

        v = d / t

        v = 2π R / T

Let's replace

        G m M / R² = m (2π R / T)² / R

        G M = R³ 4π² / T²

        T² = 4π² R³ / G M

       T² = (4π² (12 10⁴)³ / (6.67 10⁻¹¹ 4,997 10²⁰)

       T² = 6.82 10¹⁶ / 3.33 10¹⁰

       T = √ (2,048 10⁶)

       T = 1.43 10³ s

3 0
3 years ago
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Darya [45]

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One more thing—you posted this in the physics section rather than biology.

8 0
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