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kvasek [131]
3 years ago
12

Why do volcanoes form above subduction zones

Physics
1 answer:
Xelga [282]3 years ago
5 0
Volcanoes form above subduction zones because Subduction zone volcanism occurs where two plates are converging on one another. One plate containing oceanic lithosphere descends beneath the adjacent plate, thus consuming the oceanic lithosphere into the earth's mantle. This on-going process is called subduction.
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What happens to the mass and volume if density increases
nikitadnepr [17]
They all stay the same regardless
7 0
3 years ago
A particle with an initial linear momentum of 2.00 kg-m/s directed along the positive x-axis collides with a second particle, wh
ladessa [460]

Answer:

a) p₂ = 1.88 kg*m/s

   θ = 273.4 º

b)  Kf = 37% of Ko

Explanation:

a)

  • Assuming no external forces acting during the collision, total momentum must be conserved.
  • Since momentum is a vector, their components (projected along two axes perpendicular each other, x- and y- in this case) must be conserved too.
  • The initial momenta of both particles are directed one along the x-axis, and the other one along the y-axis.
  • So for the particle moving along the positive x-axis, we can write the following equations for its initial momentum:

       p_{o1x} = 2.00 kg*m/s (1)

       p_{o1y} = 0 (2)

  • We can do the same for the particle moving along the positive y-axis:

        p_{o2x} = 0 (3)

        p_{o2y} = 4.00 kg*m/s (4)

  • Now, we know the value of magnitude of the final momentum p1, and the angle that makes with the positive x-axis.
  • Applying the definition of cosine and sine of an angle, we can find the x- and y- components of the final momentum of the first particle, as follows:

       p_{f1x} = 3.00 kg*m/s * cos 45 = 2.12 kg*m/s (5)

      p_{f1y} = 3.00 kg*m/s sin 45 = 2.12 kg*m/s  (6)

  • Now, the total initial momentum, along these directions, must be equal to the total final momentum.
  • We can write the equation for the x- axis as follows:

       p_{o1x} + p_{o2x} = p_{f1x} + p_{f2x}  (7)

  • We know from (3) that p₀₂ₓ = 0, and we have the values of p₀1ₓ from (1) and pf₁ₓ from (5) so we can solve (7) for pf₂ₓ, as follows:

       p_{f2x} = p_{o1x} - p_{f1x} = 2.00kg*m*/s - 2.12 kg*m/s = -0.12 kg*m/s (8)

  • Now, we can repeat exactly the same process for the y- axis, as follows:

       p_{o1y} + p_{o2y} = p_{f1y} + p_{f2y}  (9)

  • We know from (2) that p₀1y = 0, and we have the values of p₀₂y from (4) and pf₁y from (6) so we can solve (9) for pf₂y, as follows:

       p_{f2y} = p_{o1y} - p_{f1y} = 4.00kg*m*/s - 2.12 kg*m/s = 1.88 kg*m/s (10)

  • Since we have the x- and y- components of the final momentum of  the second particle, we can find its magnitude applying the Pythagorean Theorem, as follows:

       p_{f2} = \sqrt{p_{f2x} ^{2} + p_{f2y} ^{2} }  = \sqrt{(-0.12m/s)^{2} +(1.88m/s)^{2}} = 1.88 kg*m/s (11)

  • We can find the angle that this vector makes with the positive x- axis, applying the definition of tangent of an angle, as follows:

       tg \theta = \frac{p_{2fy} }{p_{2fx} } = \frac{1.88m/s}{(-0.12m/s} = -15.7 (12)

  • The angle that we are looking for is just the arc tg of (12) which measured in a counter-clockwise direction from the positive x- axis, is just 273.4º.

b)

  • Assuming that both masses are equal each other, we find that the momenta are proportional to the speeds, so we find that the relationship from the final kinetic energy and the initial one can be expressed as follows:

       \frac{K_{f}}{K_{o} } = \frac{v_{f1}^{2} + v_{f2} ^{2}}{v_{o1}^{2} + v_{o2} ^{2} } = \frac{12.5}{20} = 0.63 (13)

  • So, the final kinetic energy has lost a 37% of the initial one.

6 0
3 years ago
What is the resultant displacement of the two vectors below?
andriy [413]

Answer:

C 3.6 cm, 56 degrees North of the East axis

Explanation:

The two vectors are perpendicular to each other, so we can find the magnitude of their resultant simply by using the Pythagorean theorem:

R=\sqrt{A^2+B^2}

where

A = 2.0 cm is the magnitude of the first vector

B = 3.0 cm is the magnitude of the second vector

Substituting,

R=\sqrt{2^2+3^2}=3.6 cm

Now we have to find the angle. If we measure the angle as North of East, the tangent of the angle is equal to the ratio between the component along North and the component along East. Therefore, in this case:

tan \theta = \frac{B}{A}=\frac{3}{2}=1.5\\\theta = tan^{-1}(1.5)=56.3 \sim 56^{\circ}

So, 56 degrees North of East.

4 0
4 years ago
5. If the average velocity of a duck is zero in a given
Ronch [10]
If the average velocity of the duck is zero, it means that the duck's location at the end of the time interval was the same as at the beginning of the interval, but says nothing about the duck's motion during that time.
For instance, the duck could have waddled around in a circle 20 times; as long as it wound up at the starting point, the displacement and average velocity is zero.
5 0
3 years ago
Which energy resource is renewable
KATRIN_1 [288]

Explanation:

solar energy is the answer

8 0
3 years ago
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