When an object is fully magnetized, all of its magnetic domains will be ?
A
Facing random directions,
B
Facing opposite directions, split down the middle of the object.
С
Facing opposite directions every-other domain
D
All facing the same direction.

Answers

Answer 1
the answer is to this question D
Answer 2
The answer would be D

Related Questions

fill in the term to complete each sentence

a wave carries___ from one place to another.
mechanical waves carry energy through___.

Answers

Energy

A material medium

A wave carries energy from one place to another. Mechanical waves carry energy through a material medium.

What is a wave?

A wave is a phenomenon that flows across a material medium without leaving any lasting mark.

Mechanical or electromagnetic waves can be used to classify them. Transverse and longitudinal are the two main forms.

A wave carries energy from one place to another. Mechanical waves carry energy through a material medium.

Hence the correct options for the fill-in-the-blanks are energy and a material medium.

To learn more about the wave refer to the link;

brainly.com/question/3004869

In the diagram, q1= +8.0 C, q2= +3.5 C, and q3 = -2.5 C. q1 to q2 is 0.10 m, q2 to q3 is 0.15 m. What is the net force on q2? Label sign for charges (+ or -). (Unit = N). pls help?

Answers

Answer:

f(t) =  28,7 [N]

Explanation: IMPORTANT NOTE: IN PROBLEM STATEMENT CHARGES ARE IN C (COULOMBS) AND IN THE DIAGRAM IN μC. WE ASSUME CHARGES ARE IN μC.

The net force on +q₂  is the sum of the force of +q₁  on +q₂ ( is a repulsion force since charges of equal sign repel each other ) and the force of -q₃ on +q₂ ( is an attraction force, opposite sign charges attract each other)

The two forces have the same direction to the right of charge q₂, we have to add them

Then

f(t) = f₁₂ + f₃₂

f₁₂ = K * ( q₁*q₂ ) / (0,1)²

q₁  = + 8 μC     then   q₁ = 8*10⁻⁶ C

q₂ =  + 3,5 μC  then  q₂ = 3,5 *10⁻⁶ C

K = 9*10⁹  [ N*m² /C²]

f₁₂ = 9*10⁹ * 8*3,5*10⁻¹²/ 1*10⁻²   [ N*m² /C²]* C*C/m²

f₁₂ = 252*10⁻¹ [N]

f₁₂ = 25,2 [N]

f₃₂ =  9*10⁹*3,5*10⁻⁶*2,5*10⁻⁶ /(0,15)²

f₃₂ =  78,75*10⁻³/ 2,25*10⁻²

f₃₂ =  35 *10⁻¹

f₃₂ =  3,5 [N]

f(t) =  28,7 [N]

Answer:

facts

Explanation:

Find the x-component of this vector: 47.3 m 39.4 ° Remember, angles are measured from the + x axis. x-component (m)

Answers

The x-component of the vector is 36.55 m.

Components of vectors

The components of a vector include both x and y components. The x-component is determined along x-axis, while y-component is determined along y-axis.

Vx = Vcosθ

Vy = Vsinθ

where;

Vx is the x-component of the vectorVy is the y-component of the vector

X-components of the vector

The x-component of the vector is calculated as follows;

x = 47.3 x cos(39.4)

x = 36.55 m

Thus, the x-component of the vector is 36.55 m.

Learn more about x-component of vector here: https://brainly.com/question/13416288

A researcher is studying traffic patterns at an intersection before an upcoming construction project. As part of this investigation, the researcher writes a scientific question, develops a procedure to observe traffic specifically at this intersection, and draws a conclusion. Which type of investigation is the researcher conducting? descriptive comparative experimental procedural

Answers

Answer:

Procedural is the answer

hope this helps

Answer: Procedural

Explanation: It states in the question that he developed a procedure! If this was something you have heard before, this is definitely the answer. Goodluck! ☺️

What is the average velocity of atoms in 2.00 mol of neon (a monatomic gas)
at 308 K? Use the equation: -mv2
2
For m, use 0.02000 kg. Remember that R = 8.31 J/(mol-K).
= ER
3
2
nRT
A. 1540 m/s
B. 876 m/s
C. 87.6 m/s
O D. 15,400 m/s

Answers

Answer:

v = 876 m/s

Explanation:

It is given that,

Number of mol of Neon is 2 mol

Temperature, T = 308 K

Mass, m = 0.02 kg

Value of R - 8.31 J/mol-K

We need to find the average velocity of atoms in 2.00 mol of neon. Neon is a monoatomic gas. Let v is the velocity. So,

[tex]\dfrac{1}{2}mv^2=\dfrac{3}{2}nRT\\\\v=\sqrt{\dfrac{3nRT}{m}} \\\\v=\sqrt{\dfrac{3\times 2\times 8.314\times 308}{0.02}} \\\\v=876.47\ m/s[/tex]

So, the correct option is (B).

I think it is 887m/s I hope this helps if not I’m really sry
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