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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A spring scale measures weight because <span>It works by Hooke's Law, which states that the force needed to extend a </span>spring<span> is proportional to the distance that </span>spring<span> is extended from its rest position. Therefore, the </span>scale<span> markings on the </span>spring<span> balance are equally spaced. A </span>spring scale<span> can</span>not measure mass<span>, only </span>weight<span>. hope that helped</span>
Answer
given,
F₁ = 15 lb
F₂ = 8 lb
θ₁ = 45°
θ₂ = 25°
Assuming the question's diagram is attached below.
now,
computing the horizontal component of the forces.
F_h = F₁ cos θ₁ - F₂ cos θ₂
F_h = 15 cos 45° - 8 cos 25°
F_h = 3.36 lb
now, vertical component of the forces
F_v = F₁ sin θ₁ + F₂ sin θ₂
F_v = 15 sin 45° + 8 sin 25°
F_v = 13.98 lb
resultant force would be equal to
![F = \sqrt{F_h^2+F_v^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Csqrt%7BF_h%5E2%2BF_v%5E2%7D)
![F = \sqrt{3.36^2+13.98^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Csqrt%7B3.36%5E2%2B13.98%5E2%7D)
F = 14.38 lb
the magnitude of resultant force is equal to 14.38 lb
direction of forces
![\theta =tan^{-1}(\dfrac{F_v}{F_h})](https://tex.z-dn.net/?f=%5Ctheta%20%3Dtan%5E%7B-1%7D%28%5Cdfrac%7BF_v%7D%7BF_h%7D%29)
![\theta =tan^{-1}(\dfrac{13.98}{3.36})](https://tex.z-dn.net/?f=%5Ctheta%20%3Dtan%5E%7B-1%7D%28%5Cdfrac%7B13.98%7D%7B3.36%7D%29)
θ = 76.48°