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2 years ago

A metal can become magnetic if the outermost electrons

Physics

1 answer:

Marizza181 [45]2 years ago

4 0

Answer:

The magnetic moment of a system measures the strength and the direction of its magnetism. The term itself usually refers to the magnetic dipole moment. Anything that is magnetic, like a bar magnet or a loop of electric current, has a magnetic moment. A magnetic moment is a vector quantity, with a magnitude and a direction. An electron has an electron magnetic dipole moment, generated by the electron's intrinsic spin property, making it an electric charge in motion. There are many different magnetic behavior including paramagnetism, diamagnetism, and ferromagnetism.

An interesting characteristic of transition metals is their ability to form magnets. Metal complexes that have unpaired electrons are magnetic. Since the last electrons reside in the d orbitals, this magnetism must be due to having unpaired d electrons. The spin of a single electron is denoted by the quantum number $m_s$ as +(1/2) or –(1/2). This spin is negated when the electron is paired with another, but creates a weak magnetic field when the electron is unpaired. More unpaired electrons increase the paramagnetic effects. The electron configuration of a transition metal (d-block) changes in a coordination compound; this is due to the repulsive forces between electrons in the ligands and electrons in the compound. Depending on the strength of the ligand, the compound may be paramagnetic or diamagnetic.Explanation:

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2) [25 pts] The bob of mass m=5 kg shown in the figure is being held by a force F. If the angle is 0⃗

Colt1911 [192]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The tension is $T = 48.255 \ N$

The time taken is $t = 0.226 \ s$

The position for maximum velocity is

S = 0

The maximum velocity is $V_{max} =0.384 \ m/s$

Explanation:

The free body for this question is shown on the second uploaded image

From the question we are told that

The mass of the bob is $m_b = 5 \ kg$

The angle is $\theta = 10^o$

The length of the string is $L = 0.5 \ m$

The tension on the string is mathematically represented as

$T = mg cos \theta$

substituting values

$T = 5 * 9.8 cos(10)$

$T = 48.255 \ N$

The motion of the bob is mathematically represented as

$S = A sin (w t + \frac{\pi }{2} )$

=> $S = A sin (wt)$

Where $w$ is the angular speed

and $\frac{\pi}{2}$ is the phase change

At initial position S = 0

So $wt = cos^{-1} (0)$

$wt = 1$

Generally $w$ can be mathematically represented as

$w = \frac{2 \pi }{T}$

Where T is the period of oscillation which i mathematically represented as

$T = 2 \pi \sqrt{\frac{L}{g} }$

$t = \frac{1}{w}$

$t = \frac{T}{2 \pi}$

$t = \sqrt{\frac{L}{g} }$

substituting values

$t = \sqrt{\frac{0.5}{9.8} }$

$t = 0.226 \ s$

Looking at the equation

$wt = 1$

We see that maximum velocity of the bob will be at S = 0

i. e $w = \frac{1}{t}$

The maximum velocity is mathematically represented as

$V_{max} = w A$

Where A is the amplitude which is mathematically represented as

$A = L sin \theta$

$V_{max} = \frac{2 \pi }{T } L sin \theta$

$V_{max} = \frac{2 \pi }{2 \pi } \sqrt{\frac{g}{L} } L sin \theta$ Recall $T = 2 \pi \sqrt{\frac{L}{g} }$

$V_{max} = \sqrt{gL} sin \theta$

substituting values

$V_{max} = \sqrt{9.8 * 0.5 } sin (10)$

$V_{max} =0.384 \ m/s$

6 0

2 years ago

A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.

Elenna [48]

Answer:

a) 6.95 m/s

b) 1.42 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s$

a) The vertical speed when it leaves the ground. is 6.95 m/s

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s$

Time taken to reach the maximum height is 0.71 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s$

Time taken to reach the ground from the maximum height is 0.71 seconds

b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds

3 0

2 years ago

Why might scientists measure the mass of object rather than the weight of an object?

Marina CMI [18]

Because they have different measurements and weight and mass and some measurements are the same

3 0

2 years ago

The ancient Egyptians used the stars to _____.

OleMash [197]

Answer:

predict annual flooding

4 0

2 years ago

Read 2 more answers

Please help homework due tomorrow,,,

Ad libitum [116K]

Cody ...

Everything on this page is solved with the SAME formula !

Distance = (speed) x (time) .

Before I get into how to solve each problem, we need to notice that
this whole sheet deals with speed, NOT velocity.

'Velocity' is speed AND THE DIRECTION OF THE MOTION.
Nothing on this page ever mentions direction, so there's no velocity
anywhere on the page.

Your teacher may not be happy if you talk about this on your homework,
but that's too bad. Just don't say "velocity" in any of your answers.
Say "speed", and if the teacher complains about that, then it's time to
let the teacher have it with both barrels.

1). Speed = (distance covered) / (time to cover the distance)

2). Speed = (distance covered) / (time to cover the distance)

3). Distance = (average speed of travel) x (time traveling at that speed)

4). Time to cover the distance = (distance) / (speed)

5). Car's speed = (distance the car covered) / (time the car took)
Sprinter speed = (distance the sprinter covered) / (time the sprinter took)

Calculate the car's speed.
Calculate the sprinter's speed.

... Look at the two speeds.
Decide which one is faster.

... Subtract the slower one from the faster one.
The difference is the answer to "by how much?" .

6). Distance = (speed) x (time spent moving at that speed)

7). Average speed = (TOTAL distance covered)
divided by
(time to cover the TOTAL distance).

8 0

2 years ago