Why do rocks on the ocean floor form a patter of magnetized strpes?

Physics

2 answers:

Citrus2011 [14]2 years ago

8 0

Answer:

The sea-floor spreading starts when the magma rises and erupts from the cracks along the mid-ocean ridges. Once the molten material solidifies, the older rock gets pushed away from the ridge and the new rock forms in the centre of the mid-ocean ridge. According to scientists, the Earth’s magnetic field is involved in this process. Since the oceanic crust contains iron, the way the rocks on the ocean floor hardened on both sides of the mid-ocean ridges are influenced by the direction of the magnetic field. As a result, it forms a pattern that resembles stripes that are alternately pointed north and south.

Or if you want a small answer:

The rock of the ocean floor contains iron. As molten material cools and hardens, the iron particles inside lines up in the direction of Earth's magnetic poles, creating a pattern of magnetized stripes.

DENIUS [597]2 years ago

4 0

Answer:

The striped magnetic pattern develops because, as the oceanic crust pulls apart, magma rises to the surface at mid-ocean ridges and spills out to create new bands of the ocean floor.

Explanation:

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In an elastic head-on collision, a 0.60 kg cart moving at 5.0 m/s [W] collides with a 0.80 kg cart moving at 2.0 m/s [E]. The co

labwork [276]

Answer:

The answer is given below

Explanation:

u is the initial velocity, v is the final velocity. Given that:

$m_1=0.6kg,u_1=-5m/s(moving \ west),m_2=0.8kg,u_2=2m/s,k=1200N/m$

The final velocity of cart 1 after collision is given as:

$v_1=(\frac{m_1-m_2}{m_1+m_2})u_1+\frac{2m_2}{m_1+m_2}u_2\\ Substituting:\\v_1=\frac{0.6-0.8}{0.6+0.8} (-5)+\frac{2*0.8}{0.6+0.8}(2)= 5/7+16/7=3\ m/s$

The final velocity of cart 2 after collision is given as:

$v_2=(\frac{m_2-m_1}{m_1+m_2})u_2+\frac{2m_1}{m_1+m_2}u_1\\ Substituting:\\v_1=\frac{0.8-0.6}{0.6+0.8} (2)+\frac{2*0.6}{0.6+0.8}(-5)= 2/7-30/7=-4\ m/s$

b) Using the law of conservation of energy:

$\frac{1}{2}m_1u_1+ \frac{1}{2}m_2u_2=\frac{1}{2}m_1v_1+\frac{1}{2}m_2v_2+\frac{1}{2}kx^2\\x=\sqrt{\frac{m_1u_1+m_2u_2-m_1v_1-m_2v_2}{k}}\\ Substituting\ gives:\\x=\sqrt{\frac{0.6*(-5)^2+0.8*2^2-(0.6*3^2)-(0.8*(-4)^2)}{1200}}=\sqrt{0}=0\ cm$