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Aleks04 [339]
3 years ago
11

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

Physics
1 answer:
Anna007 [38]3 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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Newton Metre

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A microphone is located on the line connecting two speakers that are 0.513 m apart and oscillating in phase. The microphone is 1
pantera1 [17]

Answer:

frequency 1 = 334.30 Hz

frequency 2 = 1002.92 Hz

Explanation:

Given data

speaker distance y = 0.513 m

microphone distance D = 1.80 m

to find out

lowest two frequencies

solution

we know velocity of sound is 343 m/s

so we consider point x

so at 1st speaker distance from x   = D + (y/2)

1st speaker distance from x   = 1.80 + (0.513/2) = 2.0565 m   .....1

and

at 2nd speaker distance from x   = D - (y/2)

2nd speaker distance from x   = 1.80 - (0.513/2) = 1.5435 m     .........2

so destructive interference from 1 and 2  we know

1st - 2nd = ( m + 0.5 ) wavelength

2.0565 m - 1.5435 m = (  0+ 0.5) wavelength

wavelength  = 1.026 m

so here 1st min frequency will be

frequency 1 = velocity of sound / wavelength

frequency 1 = 343 / 1.026 =334.30 Hz

and

2nd  min frequency will be

frequency 2 =

2.0565 m - 1.5435 m = (  1 + 0.5) wavelength

wavelength  = 0.342 m

frequency 2 =  velocity of sound / wavelength

frequency 2 = 343 / 0.342 = 1002.92 Hz

7 0
3 years ago
What is a Pulley why is it used​
aleksley [76]

Answer:

A simple pulley is a wheel with a rope that allows you to pull one end and have it lift whatever is on the other end. A modern, common example of this is a crane, often used in construction.

Explanation:

4 0
3 years ago
A box of mass 60 kg is at rest on a horizontal floor that has a static coefficient of friction of 0.6 and a kinetic coefficient
gavmur [86]

Answer:

a) The minimum force required to start moving the box is 352.86 N

b) i) The friction force for the box in motion is 147.025 N

ii) The acceleration of box is 4.21625 m/s²

Explanation:

The parameters of the box at rest and the floor are;

The mass of the box = 60 kg

The static coefficient friction of the floor = 0.6

The kinetic coefficient friction of the floor = 0.25

Frictional force = Normal force × Friction coefficient

For an horizontal floor and the box laying on the floor, we have;

The normal force = The weight of the box = Mass of the box × Acceleration due to gravity, g

The acceleration due to gravity, g = 9.81 m/s²

The weight of the box  = 60 × 9.81 = 588.6 N

a) The static coefficient gives the frictional force observed by the box and which must be surpassed to bring about motion

Therefore;

The minimum force required to start moving the box = The static frictional force = Weight of the box × The static coefficient of friction

The minimum force required to start moving the box = 588.1 × 0.6 = 352.86 N

The minimum force required to start moving the box = 352.86 N

b) i) When an horizontal force of 400 N is applied, the applied force is larger than the static friction force, and the box will be in motion with the kinetic coefficient of friction being the source of friction

The friction force for the box in motion = 588.1 × 0.25 = 147.025 N

ii) The force, F with the box is in motion, is given as follows;

F = Mass of box × Acceleration of box, a = Applied force - Kinematic friction force

F = 60 × a = 400 - 147.025 = 252.975 N

60 × a = 252.975 N

a = 252.975 N/(60 kg) = 4.21625 m/s²

Acceleration of box, a = 4.21625 m/s².

6 0
2 years ago
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