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4 years ago

How and where is old oceanic crust destroyed?

Physics

2 answers:

sladkih [1.3K]4 years ago

8 0

Answer:

It is destroyed in subduction zones. A Geologic process in which a tectonic plate made of dense lithospheric material melts or falls below a plate made of less-dense lithosphere at a convergent plate boundary

Explanation:

Hope this helps (:

marshall27 [118]4 years ago

7 0

<h2>Oceanic Crust - Answer</h2>

Like continental crust, however, oceanic crust is destroyed in subduction zones. The lavas are generally of two types: pillow lavas and sheet flows. Pillow lavas appear to be shaped exactly as the name implies—like large overstuffed pillows about 1 metre (3 feet) in cross section and 1 to several metres long.

Hope this helps <3

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Pluto's atmosphere. As recently observed by the New Horizons mission, the surface pressure of Pluto is about 11 microbar. The su

Komok [63]

Answer:

a) The number density is 3.623 × 10⁻³ $\frac{mol}{m^{3} }$

The mass of the atmosphere is 1.3 × 10²²Kg

b) The pressure is 10⁻²⁰ Millimeter of mercury

c) The mass mixing ratio is 0.0107

The partial pressure of ethane is 0.01114 Pa

Yes it is condensable because it boiling point is -88.5 C which is equivalent to 184.5 K i.e is adding 273 to -88.5C and the temperature of the atmosphere is 37 K.

Explanation:

The explanation is on the first and second uploaded image

4 0

3 years ago

Calculate the specific heat of a metal from the following data. A container made of the metal has a mass of 3.8 kg and contains

OLEGan [10]

Answer:

C = 771.35 J/kg°C

Explanation:

Here, e consider the conservation of energy equation. The conservation of energy principle states that:

Heat Given by Metal Piece = Heat Absorbed by Water + Heat Absorbed by Container

Since,

Heat Given or Absorbed by a material = m C ΔT

Therefore,

m₁CΔT₁ = m₂CΔT₂ + m₃C₃ΔT₃

where,

m₁ = Mass of Metal Piece = 2.3 kg

C = Specific Heat of Metal = ?

ΔT₁ = Change in temperature of metal piece = 165°C - 18°C = 147°C

m₂ = Mass of Metal Container = 3.8 kg

ΔT₂ = Change in temperature of metal piece = 18°C - 15°C = 3°C

m₃ = Mass of Water = 20 kg

C₃ = Specific Heat of Water = 4200 J/kg°C

ΔT₃ = Change in temperature of water = 18°C - 15°C = 3°C

Therefore,

(2.3 kg)(C)(147°C) = (3.8 kg)(C)(3°C) + (20 kg)(4186 J/kg°C)(3°C)

C[(2.3 kg)(147°C) - (3.8 kg)(3°C)] = 252000 J

C = 252000 J/326.7 kg°C

5 0

3 years ago

The change in momentum of an object is equal to the ____________ that acts on it.

meriva

Answer : The change in momentum of an object is equal to the impulse that acts on it.

Explanation :

Change in momentum : The change in momentum of an object is the product of the mass and the change in velocity of an object.

The formula of change in momentum is,

$\Delta p=m\times \Delta v$

Impulse : An impulse of an object is the product of the force applied on an object and the change in time. Impulse is also equivalent to the change in momentum of an object.

$J=F\times \Delta t$

Proof :

$J=F\times \Delta t\\\\J=(m\times a)\times \Delta t\\\\J=m\times (a\times \Delta t)\\\\J=m\times \Delta v=\Delta p$

Hence, the change in momentum of an object is equal to the impulse that acts on it.

3 0

3 years ago

Read 2 more answers

A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1

Archy [21]

Answer:

a) y= 3.5 10³ m, b) t = 64 s

Explanation:

a) For this exercise we use the vertical launch kinematics equation

Stage 1

y₁ = y₀ + v₀ t + ½ a t²

y₁ = 0 + 0 + ½ a₁ t²

Let's calculate

y₁ = ½ 16 10²

y₁ = 800 m

At the end of this stage it has a speed

v₁ = vo + a₁ t₁

v₁ = 0 + 16 10

v₁ = 160 m / s

Stage 2

y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²

y₂ = 800 + 150 5 + ½ 11 5²

y₂ = 1092.5 m

Speed is

v₂ = v₁ + a₂ t

v₂ = 160 + 11 5

v₂ = 215 m / s

The rocket continues to follow until the speed reaches zero (v₃ = 0)

v₃² = v₂² - 2 g y₃

0 = v₂² - 2g y₃

y₃ = v₂² / 2g

y₃ = 215²/2 9.8

y₃ = 2358.4 m

The total height is

y = y₃ + y₂

y = 2358.4 + 1092.5

y = 3450.9 m

y= 3.5 10³ m

b) Flight time is the time to go up plus the time to go down

Let's look for the time of stage 3

v₃ = v₂ - g t₃

v₃ = 0

t₃ = v₂ / g

t₃ = 215 / 9.8

t₃ = 21.94 s

The time to climb is

$t_{s}$ = t₁ + t₂ + t₃

t_{s} = 10+ 5+ 21.94

t_{s} = 36.94 s

The time to descend from the maximum height is

y = v₀ t - ½ g t²

When it starts to slow down it's zero

y = - ½ g t_{b}²

t_{b} = √-2y / g

t_{b} = √(- 2 (-3450.9) /9.8)

t_{b} = 26.54 s

Flight time is the rise time plus the descent date

t = t_{s} + t_{b}

t = 36.94 + 26.54

t =63.84 s

t = 64 s

3 0

3 years ago

The mass of the particles that a river can transport is proportional to the sixth power of the speed of the river. A certain riv

kow [346]

Answer:

1.122 m/s

Explanation:

So usually a river with a speed of 1 meters per second can transport particle that weighs:

$1^6 = 1 kg$

If the particle is twice as massive as usual, then its weights would be 1 * 2 = 2kg

This means the river must be flowing at a speed of

$2^{\frac{1}{6}} = 1.122 m/s$

5 0

3 years ago