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TEA [102]
3 years ago
9

Fe has the electron configuration [Ar]3d64s2. What is the magnitude of the orbital angular momentum for its most energetic elect

ron? √2h Võh - V12h V12h 20h
Physics
1 answer:
salantis [7]3 years ago
7 0

Answer : The magnitude of the orbital angular momentum for its most energetic electron is, \sqrt{6}\hbar

Explanation :

The formula used for orbital angular momentum is:

L=\sqrt{l(l+1)}\hbar

where,

L = orbital angular momentum

l = Azimuthal quantum number

As we are given the electronic configuration of Fe is, [Ar]3d^64s^2

Its most energetic electron will be for 3d electrons.

The value of azimuthal quantum number(l) of d orbital is, 2

That means, l = 2

Now put all the given values in the above formula, we get:

L=\sqrt{2(2+1)}\hbar

L=\sqrt{6}\hbar

Therefore, the magnitude of the orbital angular momentum for its most energetic electron is, \sqrt{6}\hbar

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From the edge of a cliff, a 0.41 kg projectile is launched with an initial kinetic energy of 1430 J. The projectile's maximum up
NemiM [27]

Answer:

v₀ₓ = 63.5 m/s

v₀y = 54.2 m/s

Explanation:

First we find the net launch velocity of projectile. For that purpose, we use the formula of kinetic energy:

K.E = (0.5)(mv₀²)

where,

K.E = initial kinetic energy of projectile = 1430 J

m = mass of projectile = 0.41 kg

v₀ = launch velocity of projectile = ?

Therefore,

1430 J = (0.5)(0.41)v₀²

v₀ = √(6975.6 m²/s²)

v₀ = 83.5 m/s

Now, we find the launching angle, by using formula for maximum height of projectile:

h = v₀² Sin²θ/2g

where,

h = height of projectile = 150 m

g = 9.8 m/s²

θ = launch angle

Therefore,

150 m = (83.5 m/s)²Sin²θ/(2)(9.8 m/s²)

Sin θ = √(0.4216)

θ = Sin⁻¹ (0.6493)

θ = 40.5°

Now, we find the components of launch velocity:

x- component = v₀ₓ = v₀Cosθ  = (83.5 m/s) Cos(40.5°)

<u>v₀ₓ = 63.5 m/s</u>

y- component = v₀y = v₀Sinθ  = (83.5 m/s) Sin(40.5°)

<u>v₀y = 54.2 m/s</u>

7 0
3 years ago
A hydraulic system for a dentist's chair is designed to be able to lift 3,112 newtons. The surface area over which this force is
In-s [12.5K]

Answer:

13.8 N

Explanation:

Pressure on the one end of the hydraulic system = Pressure on the other end

Pressure = Force / Area where Force is in Newton, area is in m²

so Force of one end (F1) / area of that end = force of the other end (F2) / area of that end

3112 / ( 707 /10000) in m² = F2 / ( 3.14 / 10000) in m²

cross multiply

44016.97 × 0.000314 = 13.82 N

5 0
3 years ago
Describe how the motion of a planet (like Mars) among the stars of the zodiac on the sky is qualitatively different from the mot
Advocard [28]

Explanation:

In the plane of the Solar System, called Ecliptic plane, Sun is at the center and all the planets including Earth are going around the Sun in their respective orbits. As seen form the Earth, Sun appear to go around it and in this process it  crosses through 12 constellations. These constellations are called as Zodiacs. Now since all the planets move in almost the same plane, they will appear in the sky to be lying near the ecliptic.

Sun's own motion is not that evident to us because of which the impact of that motion is nullified. But, motion of planet is clearly visible and it also shows in their annual round around the Sun. Here, Mars also shows retrograde motion which means it will show a back and forth motion in the sky. Because of which Mars might be visible in the same zodiac for a longer Duration as compared to the Sun.

7 0
3 years ago
After a surface interval my pressure group is D. I want to do a repetitive dive to 38 feet for 32 minutes. What will be my new p
ExtremeBDS [4]

Answer:

New pressure group is N

Explanation:

The Repetitive Dive table has to be checked, the residual nitrogen for the maximum depth of 38 feet as a "D" diver, then add that to the actual dive time to get the total nitrogen time. The total nitrogen time and the maximum depth of 38 feet would put you on a pressure group N.

5 0
3 years ago
Read 2 more answers
A football quarterback runs 15.0 m straight down the playing field in 3.00 s. He is then hit and pushed 3.00 m straight backward
algol [13]

Answer:

a) v_{1}=14.29m/s\\v_{2}=9.25m/s\\v_{3}=6.36m/s

b) v=+9.97m/s

Explanation:

From the exercise we know that

x_{1} =15m, t_{1}=3s

x_{2} =-3m, t_{1}=1.74s

x_{3} =29m, t_{3}=5.20s

From dynamics we know that the formula for average velocity is:

v=\frac{x_{2}-x_{1}  }{t_{2}-x_{1}  }

a) For the three intervals:

v_{1}=\frac{x_{2}-x_{1}  }{t_{2}-t_{1}  }=\frac{(-3-15)m}{(1.74-3)s}=14.29m/s

v_{2}=\frac{x_{3}-x_{2}  }{t_{3}-t_{2}  }=\frac{(29-(-3))m}{(5.20-1.74)s}=9.25m/s

v_{3}=\frac{x_{3}-x_{1}  }{t_{3}-t_{1}  }=\frac{(29-15)m}{(5.20-3)s}=6.36m/s

b) The average velocity for the entire motion can be calculate by the following formula:

v=\frac{v_{1}+v_{2}+v_{3}   }{n} =\frac{(14.29+9.25+6.36)m/s}{3}=+9.97m/s

8 0
3 years ago
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