Answer:
I think A
Explanation:
because it dosn't have enough tools
The magnet (south pole of the magnet) has magnetized the right side of the block.
<h3>
Direction of electric field in the magnetic material</h3>
The direction of electric field of the atom of the magnetic material is unpolarized.
From the diagram in the image, the right hand side of the magnetic material is being attracted to south pole of the magnet.
Thus, we can conclude that, the magnet has magnetized the right side of the block.
Learn more about magnetic material here: brainly.com/question/22074447
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Complete Question
The complete question is shown on the first uploaded image
Compare the percent error between your value and the accepted value of 1.26 times 10-6 T · m/A. (Use the accepted value given to three significant figures in your calculation.). 100% % error = %
Answer:
The permeability of free space is 
The percentage error is % error = 5.25%
Explanation:
From the question we are told that
The slope is
, and
Generally 
So 
From the relation given in the question
![\mu_0 = \frac{2R}{N} [\frac{B}{I} ]](https://tex.z-dn.net/?f=%5Cmu_0%20%3D%20%5Cfrac%7B2R%7D%7BN%7D%20%5B%5Cfrac%7BB%7D%7BI%7D%20%5D)
Where R is the radius of the coil 
N is the number of loops of the coil = 10
Now from the question we are told that

substituting into the equation above

Substituting values


Generally the % error is mathematically represented as
%
Given that the accepted value is
Hence substituting values
%

Answer:
The electric potential in volts is 1.618 x 10⁻¹⁷ V
Explanation:
The electric potential, in volts, at point P, can be calculated as follow;
Electric potential is the work done in moving a unit positive charge from infinity to a particular point in the electric field.
Thus, the work done in this process in moving the charge to point p is 101eV.
Convert this Volts = 101 × 1.602 x 10⁻¹⁹ V
= 1.618 x 10⁻¹⁷ V
Therefore, the electric potential in volts is 1.618 x 10⁻¹⁷ V
Radar waves are the waves with the lowest energy.