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Zigmanuir [339]
4 years ago
11

How would you use Tectonic plate and hot spot in the same sentence

Physics
2 answers:
Sonbull [250]4 years ago
7 0
It's kinda long but...
A tectonic setting where volcanic action occurs is called <span>a </span>hot-spot (intraplate<span>), which describes volcanic activity that occurs </span>within tectonic plates<span> and is generally NOT related to plate boundaries and plate movements.
</span>Hope this helps!!:)
jok3333 [9.3K]4 years ago
6 0
Volcanic hot spot is where volcanoes are <span>fed by underlying </span>mantle<span> that is anomalously hot and such volcanoes are different in origin from other volcanoes formed over </span>subduction<span> zones where tectoic plates run into each other.</span>
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The diagram shows a spectrometer being used to analyze a sample of a solution.
fredd [130]

Answer

B. X

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3 years ago
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PolarNik [594]

Answer: The plasma membrane is called a selectively permeable membrane as it permits the movement of only certain molecules in and out of the cells. Not all molecules are free to diffuse. If plasma membrane ruptures or breaks down then molecules of some substances will freely move in and out of the cells.

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3 years ago
Two ranger stations are on an east-west line 110 mi apart. A forest fire is located on a bearing of N 42º E from the western sta
olchik [2.2K]

Answer:

Explanation:

AB = 110 miles

Let the distance of the western station from fire is d.

As according to the diagram, use Sine law

\frac{d}{Sin 15}=\frac{110}{Sin 133}

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Use the information below to answer questions
Ulleksa [173]

Answer:

The charges are q₁  = 2 × 10⁻⁸ C and  q₂ = 3 × 10⁻⁸ C

Explanation:

Here is the complete question

Two identical tiny balls have charge q1 and q2. The repulsive force one exerts on the other when they are 20cm apart is 1.35 X 10-4 N. after the balls are touched together and then represented once again to 20cm, now the repulsive force is found to be 1.40 X 10-4 N. find the charges q1 and q2.

Solution

The force F = 1.35 × 10⁻⁴ N when the charges are separated a distance of r = 20 cm = 0.2 m is given by

F = kq₁q₂/r₁²

q₁q₂ = Fr₁²/k

q₁q₂ = 1.35 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.054/9 × 10⁻¹³ C² = 0.006 × 10⁻¹³ C² = 6 × 10⁻¹⁶ C²

q₁q₂ = 6 × 10⁻¹⁶ C² (1)

When the charges are brought together, the charge is now q = (q₁ + q₂)/2

The new repulsive force F = 1.406 × 10⁻⁴ N  at a distance of r₂ = 20 cm = 0.2 m is then

F₂ = kq²/r₂²

q² = F₂r₂²/k = 1.406 × 10⁻⁴ N × (0.2 m)²/9 × 10⁹ Nm²/C² = 0.00625 × 10⁻¹³ C² = 6.25 × 10⁻¹⁶ C²

q² = 6.25 × 10⁻¹⁶ C²

q = √(6.25 × 10⁻¹⁶) C

q = 2.5 × 10⁻⁸ C

(q₁ + q₂)/2 =  2.5 × 10⁻⁸ C

(q₁ + q₂) = 2 × 2.5 × 10⁻⁸ C

q₁ + q₂ = 5 × 10⁻⁸ C (2)

q₁  = 5 × 10⁻⁸ C - q₂  (3)

Substituting equation (3) into (1), we have

(5 × 10⁻⁸ C - q₂)q₂ = 6 × 10⁻¹⁶ C²

Expanding the bracket, we have

(5 × 10⁻⁸ C)q₂ - q₂² = 6 × 10⁻¹⁶ C²

So, q₂² - (5 × 10⁻⁸ C)q₂ + 6 × 10⁻¹⁶ C² = 0

Using the quadratic formula to find q₂

q_{2} = \frac{-(-5 X 10^{-8} )+/- \sqrt{(-5 X 10^{-8} )^{2} - 4X1X6 X 10^{-16} } }{2X1}\\  = \frac{5 X 10^{-8} )+/- \sqrt{25 X 10^{-16}  - 24 X 10^{-16} } }{2}\\= \frac{5 X 10^{-8} )+/- \sqrt{1 X 10^{-16} } }{2}\\= \frac{5 X 10^{-8} )+/- 1 X 10^{-8} }{2}\\= \frac{5 X 10^{-8} + 1 X 10^{-8} }{2} or \frac{5 X 10^{-8}  - 1 X 10^{-8} }{2}\\= \frac{6 X 10^{-8} }{2} or \frac{4 X 10^{-8}}{2}\\= 3 X 10^{-8} C or 2 X 10^{-8} C

q₁  = 5 × 10⁻⁸ C - q₂

q₁  = 5 × 10⁻⁸ C - 3 × 10⁻⁸ C or 5 × 10⁻⁸ C - 2 × 10⁻⁸ C

q₁  = 2 × 10⁻⁸ C or 3 × 10⁻⁸ C

So the charges are q₁  = 2 × 10⁻⁸ C and  q₂ = 3 × 10⁻⁸ C

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4 years ago
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Naily [24]

Answer:

1.) A - 1,2,1,2

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3.) C

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4.) A, B and C

Explanation:

1.) SO2 + ___ H2 ----------> _____ S + _______H2O

Looking at above equation, we have one molecule of surfur at the right hand side (RHS), and one molecule of surfur at the left hand side (LHS).

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But oxygen is not balanced. We have two atoms of oxygen at the RHS while having one at the LHS.

So let's make oxygen 2 atoms at LHS by adding 2 to water molecules and hydrogen molecules at the RHS.

SO2 + ___ 2H2 ----------> _____ S + _______2H2O

The correct answer is A - 1,2,1,2

2.) Both magnetic force and gravitational force obey inverse square law with distance. They are not directly proportional but inversely proportional to the square distance. B is the correct answer because Both forces can act without objects touching.

3.) Light waves are transverse waves which can travel through a vacuum without a medium while sound waves are longitudinal waves which cannot travel through vacuum without a medium. But both can be modeled as waves having amplitude, frequency and wave lengths.

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The potential energy could be increasing, like a ball thrown into the air.

The kinetic energy could be lost to friction or air resistance.

The ball could be returning to its natural resting state.

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3 years ago
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