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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.
Physics
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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1 eV

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At the same instant that a 0.50 kg ball is dropped from 25 m above Earth, a second ball, with a mass of .25 kg, is thrown straig
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Answer:

A. 7.1m

B. 3.55m/s

C. 1.775m/s^2

Explanation:

First step is to identify given parameters;

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Ball 2: m₂ = 0.25kg, u = 15m/s, t = 2seconds

<u>Second step:</u> we determine the y-coordinate of ball 1 after 2 seconds, using the equation of motion under gravity as shown below;  

y = ut - \frac{gt^2}{2}

y_{1} = 0 X 2 - \frac{9.8 X2^2}{2}

y_{1} = -19.6m

Recall, that the ball was thrown from a height of 25m, total y-coordinate of ball 1 after 2 seconds becomes 25m +(-19.6m)

[tex]y_{1}  = 5.4m[/tex]

<u>Third step</u>: we determine the y-coordinate of ball 2 after 2 seconds

y_{2} = 15 X 2 - \frac{9.8 X2^2}{2}

y_{2} = 10.4m

<u>Fourth step: </u>we determine the y-component of the center mass of the two balls

y = \frac{m_{1}y_{1} +m_{2}y_{2}}{m_{1} +m_{2} }

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Velocity = \frac{distance of center mass of the two balls (y)}{time}

velocity = \frac{7.1 m}{2 s}

velocity = 3.55m/s

<u>Sixth step:</u> we solve C part of the question; acceleration of the center mass of the two balls

acceleration = \frac{velocity}{time}

acceleration = \frac{3.55}{2}

acceleration = 1.775 m/s^2

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Answer:

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Explanation:

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