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3 years ago

3. According to Hund's rule, what's the expected magnetic behavior of vanadium (V)?

Physics

1 answer:

ivanzaharov [21]3 years ago

6 0

Answer:

Diamagnetic

Explanation:

Hunds rule states that electrons occupy each orbital singly first before pairing takes place in degenerate orbitals. This implies that the most stable arrangement of electrons in an orbital is one in which there is the greatest number of parallel spins(unpaired electrons).

For vanadium V ion, there are 18 electrons which will be arranged as follows;

1s2 2s2 2p6 3s2 3p6.

All the electrons present are spin paired hence the ion is expected to be diamagnetic.

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Which sequence shows electromagnetic waves arranged in a decreasing order of their wavelengths?

Hunter-Best [27]

Answer: Gamma rays, x-rays, ultraviolet rays, visible light, and infrared rays.

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2 years ago

Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.300 m and carries a current

Marrrta [24]

Answer:

Explanation:

Since the wires attract each other , the direction of current will be same in both the wires .

Let I be current in wire which is along x - axis

force of attraction per unit length between the two current carrying wire is given by

$\frac{\mu_0}{4\pi}$ x $\frac{2 I_1\times I_2}{d}$

where I₁ and I₂ are currents in the wires and d is distance between the two

Putting the given values

285 x 10⁻⁶ = 10⁻⁷ x $\frac{2\times25.5\times I_2}{.3}$

I₂ = 16.76 A

Current in the wire along x axis is 16.76 A

To find point where magnetic field is zero due the these wires

The point will lie between the two wires as current is in the same direction.

Let at y = y , the neutral point lies

k 2 x $\frac{16.76}{y}$ = k 2 x $\frac{25.5}{.3-y}$

25.5y = 16.76 x .3 - 16.76y

42.26 y = 5.028

y = .119

= .12 m

3 0

3 years ago

A basketball is held over head at a height of 2.4 m. The ball is lobbed to a teammate at 8 m/s at an angle of 40'. If the ball i

cupoosta [38]

Explanation:

since both the teammates are of the same height, their height won't matter. Because now the basketball won't cover any vertical distance.

We have to calculate its range the horizontal distance covered by it when tossed from one teammate to the other.

range can be calculated by the formula :-

$\boxed{\mathfrak{range = \frac{ u {}^{2} \sin 2\theta }{g} }}$

u is the velocity during its take off and $\theta$ is the angle at which its thrown

Given that

u = 8m/ s
$\theta$ = 40°

calculating range using the above formula

$= \frac{ {8}^{2} \sin2(40) }{10}$

$= \frac{64 \times \sin(80) }{10}$

value of sin 80 = 0. 985

$= \frac{64 \times 0.985}{10}$

$= \frac{63.027}{10}$

$= 6.3027$

Hence,

$\mathfrak { \blue{the \: teammate \: is \: \red{\underline{6.3027 \: meters} }\: away } }$

7 0

3 years ago

A 10 kilogram lump of rock weighs 16N on the Moon.

qwelly [4]

Answer:

Explanation:

it's the only one that makes sense

4 0

2 years ago

Read 2 more answers

•• CP Two blocks connected by a light horizontal rope sit at rest on a horizontal, frictionless surface. Block AA has mass 15.0

Firdavs [7]

Answer:

(a) T= 38.4 N

(b) m= 26.67 kg

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Kinematics

d= v₀t+ (1/2)*a*t² (Formula 2)

d:displacement in meters (m)

t : time in seconds (s)

v₀: initial speed in m/s

vf: final speed in m/s

a: acceleration in m/s²

v₀=0, d=18 m , t=5 s

We apply the formula 2 to calculate the accelerations of the blocks:

d= v₀t+ (1/2)*a*t²

18= 0+ (1/2)*a*(5)²

a= (2*18) / ( 25) = 1.44 m/s² to the right

We apply Newton's second law to the block A

∑Fx = m*ax

60-T = 15*1.44

60 - 15*1.44 = T

T = 38.4 N

We apply Newton's second law to the block B

∑Fx = m*ax

T = m*ax

38.4 = m*1.44

m= (38.4) / (1.44)

m = 26.67 kg

7 0

3 years ago