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Len [333]
1 year ago
10

Substances that move to the stronger parts of a magnetic field are termed ______ substances; the atomic feature responsible for

this property is ______ in atoms.
Physics
1 answer:
Arisa [49]1 year ago
3 0

Substances that move to the stronger parts of a magnetic field are termed paramagnetic substances; the atomic feature responsible for this property is presence of unpaired electrons in atoms.

<h3>What is a paramagnetic substance?</h3>

A paramagnetic substance is the substance that possess unpaired electrons that are heavily attracted in a magnetic field.

A magnetic field is defined as the field that exists around a magnet that produces a field of force.

Examples of paramagnetic substance include the following:

  • aluminum,
  • gold,
  • copper.
  • Chromium, and
  • Manganese.

These substances are known as paramagnetic substances because they possess a high number of unpaired electrons.

Other properties of a paramagnetic substance include the following:

  • They have a permanent dipole moment or permanent magnetic moment.
  • They are weakly magnetized in the direction of the magnetizing field.
  • They usually have constant relative permeability (μr) slightly greater than 1.

Therefore, Substances that move to the stronger parts of a magnetic field are termed paramagnetic substances; the atomic feature responsible for this property is presence of unpaired electrons in atoms.

Learn more about magnets here:

brainly.com/question/26171648

#SPJ1

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a) 0.9995c

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c) 91670 MeV

Explanation:

(a) The speed of approach is given by the formula:

u=\frac{v_1+v_2}{1+\frac{v_1v_2}{c}}=\frac{2(0.9898c)}{1+\frac{(0.9898)^2c^2}{c^2}}=0.99995c

(b) the kinetic energy is given by:

E_k=m_0c^2[\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1]

by replacing c=3*10^8m/s, m_0=1.67*10^{-27}kg we obtain:

E_k=5641MeV

(c) in the rest frame of the other proton we have:

E_k=m_0c^2[\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}-1]

by replacing we get

E_k=91670MeV

hope this helps!!

7 0
3 years ago
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A force of 10 N causes a spring to extend by 20 mm. Find: a) the spring constant of the spring in N/m​
Mars2501 [29]

Answer:

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6 0
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At a given instant the bottom A of the ladder has an acceleration aA = 4 f t/s2 and velocity vA = 6 f t/s, both acting to the le
Nana76 [90]

Answer:

Acceleration=24.9ft^2/s^2

Angular acceleration=1.47rads/s

Explanation:

Note before the ladder is inclined at 30° to the horizontal with a length of 16ft

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acceleration Ab=Aa +(Ab/a)+(Ab/a)t

4+0.75^2*16+a*16

0=0.75^2*16cos30°-a*16sin30°---1

Ab=0+0.75^2sin30°+a*16cos30°----2

Solving equation 1

(0.75^2*16cos30/16sin30)=angular acceleration=a=1.47rad/s

Also from equation 2

Ab=0.75^2*16sin30+1.47*16cos30=24.9ft^2/s^2

6 0
3 years ago
two children of masses m1 = m2 = 20 kg , ride on the perimeter of a small merry-go- round . the merry -go-round us a disk of mas
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c

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