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2 years ago

Which type of plate boundary or zone would be most likely to lead to above-ground volcanic activity?

Physics

1 answer:

Anna007 [38]2 years ago

6 0

Answer:

Convergent plate boundary

Explanation:

The convergent plate boundary refers to the type of boundary where two plates move towards each other. Due to this type of motion, there forms a subduction zone, where the denser plate subducts below the lighter plate. This zone of subduction is commonly identified by the presence of a deep and narrow V-shaped depression which is commonly known as the oceanic trench.

When the subducting plate enters into the region of the asthenosphere, the rocks melt and mix with the magma. This magma is then pushed upward due to the force exerted by the convection current that forms in the mantle, and further reaches the over-riding plate and eventually give rise to the formation of volcanoes and volcanic/island arcs.

Thus, this type of plate boundary is responsible for the formation of above-ground volcanic activities.

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lana66690 [7]

Answer:

Height of cliff = S = 20 m (Approx)

Explanation:

Given:

Initial velocity = 8 m/s

Distance s = 16 m

Starting acceleration (a) = 0

Computation:

s = ut + 1/2a(t)²

16 = 8t

t = 2 sec

Height of cliff = S

Gravitational acceleration = 10 m/s

S = 1/2a(t)²

S = 1/2(10)(2)²

Height of cliff = S = 20 m (Approx)

3 0

3 years ago

The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 459 nm. What is t

spin [16.1K]

Answer:

2.7067 eV

Explanation:

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

c = Speed of light = $3\times 10^8\ m/s$

$\lambda_0$ = Threshold wavelength = 459 nm

Work function is given by

$W_0=\frac{hc}{\lambda_0}\\\Rightarrow W_0=\frac{6.626\times 10^{-34}\times 3\times 10^8}{459\times 10^{-9}}\\\Rightarrow W_0=4.33072\times 10^{-19}\ J$

Converting to eV

$1\ J=\frac{1}{1.6\times 10^{-19}}\ eV$

$4.33072\times 10^{-19}\ J=4.33072\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}\ eV=2.7067\ eV$

The work function W0 of this metal is 2.7067 eV

4 0

3 years ago

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alexandr402 [8]

Answer:

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Explanation:

574.780616 m6 kg3 s-6 K-3 mol-3

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3 years ago

A rocket on Earth experiences an upward applied force from its thrusters. As a result of this force, the rocket accelerates upwa

gayaneshka [121]

Answer:

F=m(11.8m/s²)

For example, if m=10,000kg, F=118,000N.

Explanation:

There are only two vertical forces acting on the rocket: the force applied from its thrusters F, and its weight mg. So, we can write the equation of motion of the rocket as:

$F-mg=ma$

Solving for the force F, we obtain that:

$F=ma+mg=m(a+g)$

Since we know the values for a (2m/s²) and g (9.8m/s²), we have that:

$F= m(2m/s^{2}+9.8m/s^{2})\\\\F=m(11.8m/s^{2})$

From this relationship, we can calculate some possible values for F and m. For example, if m=10,000kg, we can obtain F:

$F=(10,000kg)(11.8m/s^{2})\\\\F=118,000N$

In this case, the force from the rocket's thrusters is equal to 118,000N.

5 0

3 years ago

A ball is thrown straight up. At the top of its path its acceleration is

timurjin [86]

C. Acceleration is the rate of change of velocity. So at the top of the path, while the velocity is zero, the CONSTANT GRAVITATIONAL ACCELERATION is about 10 m/s^2 (9.8)

6 0

3 years ago