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3 years ago

The contact between lithospheric plates is called a

Physics

1 answer:

Alexxx [7]3 years ago

5 0

Answer:

The contact between lithospheric plates is called a. plate boundary. The center of a mid-ocean ridge is where. new oceanic lithosphere is being created.

Explanation:

The Earth’s lithosphere, which includes the crust and upper mantle, is made up of a series of pieces, or tectonic plates, that move slowly over time. A divergent boundary occurs when two tectonic plates move away from each other. Along these boundaries, earthquakes are common and magma (molten rock) rises from the Earth’s mantle to the surface, solidifying to create new oceanic crust. The Mid-Atlantic Ridge is an example of divergent plate boundaries. When two plates come together, it is known as a convergent boundary. The impact of the colliding plates can cause the edges of one or both plates to buckle up into a mountain ranges or one of the plates may bend down into a deep seafloor trench. A chain of volcanoes often forms parallel to convergent plate boundaries and powerful earthquakes are common along these boundaries. The Pacific Ring of Fire is an example of a convergent plate boundary. At convergent plate boundaries, oceanic crust is often forced down into the mantle where it begins to melt. Magma rises into and through the other plate, solidifying into granite, the rock that makes up the continents. Thus, at convergent boundaries, continental crust is created and oceanic crust is destroyed. Two plates sliding past each other forms a transform plate boundary. One of the most famous transform plate boundaries occurs at the San Andreas fault zone, which extends underwater. Natural or human-made structures that cross a transform boundary are offset—split into pieces and carried in opposite directions. Rocks that line the boundary are pulverized as the plates grind along, creating a linear fault valley or undersea canyon. Earthquakes are common along these faults. In contrast to convergent and divergent boundaries, crust is cracked and broken at transform margins, but is not created or destroyed.

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State one effect of increase in heat for a temperature range of 50°C to 100°C on Plastic

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Answer:

why is fraction called a necessary evil

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3 years ago

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.85 kg and length L = 5.76 m to a uniform sphere with

Ket [755]

Answer:

Part a)

$I = 1879.7 kg m^2$

Part b)

$\alpha = 0.70 rad/s^2$

Part c)

$I = 153.8 kg m^2$

Part 4)

angular acceleration will be ZERO

Part 5)

$I = 345.6 kg m^2$

Explanation:

Part a)

Moment of inertia of the system about left end of the rod is given as

$I = \frac{m_r L^2}{3} + (\frac{2}{5} m_s R^2 + m_s(R + L)^2)$

So we have

$I = \frac{m_r(4R)^2}{3} + (\frac{2}{5}(5m_r) R^2 + (5m_r)(R + 4R)^2)$

$I = \frac{16}{3}m_r R^2 + (2m_r R^2 + 125 m_rR^2)$

$I = (\frac{16}{3} + 127)m_r R^2$

$I = (\frac{16}{3} + 127)(6.85)(1.44)^2$

$I = 1879.7 kg m^2$

Part b)

If force is applied to the mid point of the rod

so the torque on the rod is given as

$\tau = F\frac{L}{2}$

$\tau = 460(2R)$

$\tau = 460 \times 2 \times 1.44$

$\tau = 1324.8 Nm$

now angular acceleration is given as

$\alpha = \frac{\tau}{I}$

$\alpha = \frac{1324.8}{1879.7}$

$\alpha = 0.70 rad/s^2$

Part c)

position of center of mass of rod and sphere is given from the center of the sphere as

$x = \frac{m_r}{m_r + m_s}(\frac{L}{2} + R)$

$x = \frac{m_r}{6 m_r}(3R) = \frac{R}{2}$

so moment of inertia about this position is given as

$I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + \frac{R}{2})^2 + (\frac{2}{5} m_s R^2 + m_s(\frac{R}{2})^2)$

so we have

$I = \frac{m_r (16R^2)}{12} + m_r(\frac{5R}{2})^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(\frac{R^2}{4})$

$I = m_r R^2(\frac{16}{12} + \frac{25}{4} + 2 + \frac{5}{4})$

$I = 6.85(1.44)^2\times 10.83$

$I = 153.8 kg m^2$

Part 4)

If force is applied parallel to the length of rod

then we have

$\tau = \vec r \times \vec F$

$\tau = 0$

so angular acceleration will be ZERO

Part 5)

moment of inertia about right edge of the sphere is given as

$I = \frac{m_r L^2}{12} + m_r(\frac{L}{2} + 2R)^2 + (\frac{2}{5} m_s R^2 + m_s(R)^2)$

so we have

$I = \frac{m_r (16R^2)}{12} + m_r(4R)^2 + \frac{2}{5}(5m_r)R^2 + (5m_r)(R^2)$

$I = m_r R^2(\frac{16}{12} + 16 + 2 + 5)$

$I = 6.85(1.44)^2\times 24.33$

$I = 345.6 kg m^2$

6 0

3 years ago

Sound waves are_________Waves.

Anastasy [175]

Sound waves are mechanical(?) waves?

4 0

3 years ago

Read 2 more answers

Why do rivers usually have water in them, even during a drought?

Kipish [7]

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7 0

4 years ago

Light enters an equilateral prism with an incident angle of 35Â° to the normal of the surface. Calculate the angle at which the

julia-pushkina [17]

Answer:

65.9°

Explanation:

When light goes through air to glass

angle of incidence, i = 35°

refractive index, n = 1.5

Let r be the angle of refraction

Use Snell's law

$n=\frac{Sini}{Sinr}$

$1.5=\frac{Sin35}{Sinr}$

Sin r = 0.382

r = 22.5°

Now the ray is incident on the glass surface.

A = r + r'

Where, r' be the angle of incidence at other surface

r' = 60° - 22.5° = 37.5°

Now use Snell's law at other surface

$\frac{1}{n}=\frac{Sinr'}{Sini'}$

Where, i' be the angle at which the light exit from other surface.

$\frac{1}{1.5}=\frac{Sin37.5'}{Sini'}$

Sin i' = 0.913

i' = 65.9°

4 0

3 years ago