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sergiy2304 [10]

3 years ago

14

Low-viscosity lava a. is most often a cool-temperature lava. b. could logically build a composite volcano. c. has low silica con

tent. d. indicates an area that has high potential for explosive eruptions.

1 answer:

OverLord2011 [107]3 years ago

8 0

<u>Full question:</u>

Identify the False statement. Lava of low viscosity

a. is most often a cool-temperature lava.

b. could logically build a composite volcano.

c. has low silica content.

d. indicates an area that has high potential for explosive eruptions.

<u>Answer:</u>

is most often a cool-temperature lava is the false statement about Lava of low viscosity

<u>Explanation:</u>

When lava begets low viscosity, it can move quite smoothly across extended lengths. This produces the perfect outpourings of lava, byways, plashes, and sprays. You can likewise notice globules of lava-filled amidst volcanic gasses that burble and rise on the facade of the lava.

And extra time, volcanoes produced from low lava viscosity are extensive and have a depthless incline certain are perceived as shield volcanoes. Anywhere a volcano provides low viscosity, runny, lava it flattens faraway from the origin producing a volcano with moderate inclines.

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Temperature is a measure of the average ____________ energy of an object's particles. light mechanical potential kinetic

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Temperature is a measure of the average kinetic energy of the particles of a substance.
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What is the name for the bunched up region of<br> air particles in the sound wave?

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A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .

(The other root is negative and is thus invalid.)

Hence, the maximum compression of this spring would be approximately $0.560\; {\rm m}$ .

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3 years ago

Write an expression for the magnitude of charge moved, Q, in terms of N and the fundamental charge e

NeTakaya

We have that for the Question "Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be said its equation is

$Q=\frac{E}{Nr^2}$

From the question we are told

Write an expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e

<h3>An Expression for the <em>magnitude </em>of charge moved</h3>

Generally the equation for the <em>magnitude </em>of charge moved, Q is mathematically given as

$Q=\frac{E}{Nr^2}$

Therefore

An expression for the <em>magnitude </em>of charge moved, Q, in terms of N and the fundamental charge e" it can be

$Q=\frac{E}{Nr^2}$

For more information on this visit

brainly.com/question/16517842

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2 years ago