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IRINA_888 [86]
3 years ago
12

What new tool did geologists use to estimate the changing flow rates of lava over the duration of this eruption? Hint: The answe

r is mentioned in the Kīlauea talk!a. GPS receiversb. Photographs from dronesc. Fluid dynamic calculationsd. Thermal sensors
Physics
1 answer:
Vladimir [108]3 years ago
4 0

Answer:

Option b, pothographs from drones.

Explanation:

the USGS (U.S. Geological Survey) decided to make photographic captures from drones to the volcanic surfaces, which allowed through observations to understand things like the characteristics of the lava, the height of the volcanic plumes (among others).

Podemos ver en el siguiente enlace un ejemplo de fotografía tomada desde un dron al Kilauea.

https://www.usgs.gov/media/images/k-lauea-volcano-drone-over-lava-channel

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Elements on the periodic table are grouped by their
vodka [1.7K]
Similar chemical behavior. All the members of a group of elements have the same number of valence electrons and similar chemical properties.
5 0
4 years ago
Read 2 more answers
A 10,844.0 kg truck is initially at rest. The truck then accelerates across a level road and reaches a constant speed. It takes
lara [203]

Answer:

1,634.1 W

Explanation:

Power = work / time

P = 41,505.4 J / 25.4 s

P = 1,634.1 W

7 0
3 years ago
Which table is it I’ll mark brainlest
alexira [117]

Answer:

we need the graph to answer the question.

5 0
3 years ago
What is the heat needed to raise the temperature of 24.7 kg silver from 14.0 degrees celsius to 30.0 degrees celsius? specific h
Citrus2011 [14]
The amount of heat needed to increase the temperature of a substance by \Delta T is given by
Q=m C_s \Delta T
where
m is the mass of the substance
C_s is its specific heat capacity
\Delta T is the increase of temperature

The sample of silver of our problem has a mass of m=24.7 kg. Its specific heat capacity is C_s = 236 J/g^{\circ}C and the increase in temperature is
\Delta T=30.0^{\circ}-14.0^{\circ}C=16.0^{\circ}C

Therefore, the amount of heat needed is
Q=mC_s \Delta T=(24.7 kg)(236 J/g^{\circ}C)(16.0^{\circ}C)=9.32 \cdot 10^4 J
8 0
4 years ago
awhite billiard ball with mass mw = 1.47 kg is moving directly to the right with a speed of v = 3.01 m/s and collides elasticall
pochemuha

Answer:

speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

initial kinetic energy = final kinetic energy

KE = 6.66 J

Explanation:

Since there is no external force on the system of two balls so here total momentum of two balls initially must be equal to the total momentum of two balls after collision

So we will have

momentum conservation along x direction

m_1v_{1i} + m_2v_{2i} = m_1v_{1x} + m_2v_{2x}

now plug in all values in it

1.47 \times 3.01 + 0 = 1.47 v_1cos68 + 1.47 v_2cos22

so we have

3.01 = 0.375v_1 + 0.927v_2

similarly in Y direction we have

m_1v_{1i} + m_2v_{2i} = m_1v_{1y} + m_2v_{2y}

now plug in all values in it

0 + 0 = 1.47 v_1sin68 - 1.47 v_2sin22

so we have

0 = 0.927v_1 - 0.375v_2

v_2 = 2.47 v_1

now from 1st equation we have

3.01 = 0.375 v_1 + 0.927(2.47 v_1)

v_1 = 1.13 m/s

v_2 = 2.78 m/s

so speed of white ball is 1.13 m/s and speed of black ball is 2.78 m/s

Also we know that since this is an elastic collision so here kinetic energy is always conserved to

initial kinetic energy = final kinetic energy

KE = \frac{1}{2}(1.47)(3.01^2)

KE = 6.66 J

5 0
4 years ago
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