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Alexxx [7]
4 years ago
8

Earthquakes are not easy to predict, but volcanic eruptions have more warning signals and instruments to measure those signals,

therefore making them somewhat easier to forecast. The difficulty with predicting volcanic eruptions is the___.
A. Height of the volcanoes
B. Number of volcanoes
C. Shortage of geologists
D. Cost of monitoring
Chemistry
2 answers:
r-ruslan [8.4K]4 years ago
8 0
Earthquakes may take place every day near a volcano. But before an eruption, the number and size of earthquakes<span> increases. This is the result of magma pushing upward into the magma chamber. This motion causes stresses on neighboring rock to build up. Eventually the ground shakes. A continuous string of earthquakes may indicate that a volcano is about to erupt. Scientists use seismographs to record the length and strength of each earthquake.</span>
muminat4 years ago
8 0

Answer:

D. Cost of monitoring

Explanation:

Despite the great concern and commitment of volcanologists, it is still not possible to predict exactly when a volcanic eruption will occur. However, some signs may be indicative of some upcoming activity.  Before an eruption, magma focuses on areas below the volcano, called reservoirs, where it moves producing vibrations, ie small earthquakes. This same movement can cause the slopes of the volcanic cone to collapse. As magma approaches the surface, it releases gases that can be detected in regions near the volcano.

The analysis of these gases allows to verify if there is variation in their quantity and composition. The movement of the magma, however, may not result in eruption. Instead, it can cool and solidify on the subsurface.

Since volcanoes typically come into operation at intervals of hundreds to thousands of years, they are not continuously monitored, even because this practice is costly (which makes monitoring difficult) and the same pattern is not always repeated in different eruptions.

However, when the volcano emits precursor clues, a major catastrophe can be prevented.

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Substance A undergoes a first order reaction A  B with a half- life of 20 min at 25 °C. If the initial concentration of A ([A]◦
nydimaria [60]

The concentration of A will be <em>0.34 mol·L⁻¹</em> after 60 min.

In a first-order reaction, the formula for the amount remaining after <em>n</em> half-lives is

\text{[A]} = \frac{\text{[A]}_{0}}{2^{n}}\\

If t_{\frac{1}{2}} = \text{20 min}\\

n = \frac{t}{t_{\frac{1}{2}}} = \frac{\text{60 min}}{\text{20 min}}= \text{3.0}

∴ \text{[A]} = \frac{\text{2.7 mol/L}}{2^{3.0}} = \frac{\text{2.7 mol/L}}{8.0} = \textbf{0.34 mol/L}

6 0
3 years ago
Describe what conditions exist in water molecules to make them dipolar.
lesantik [10]
<span>The bent geometry of the water molecule gives a slight overall negative charge to the oxygen side of the molecule and a slight overall positive charge to the hydrogen side of the molecule. This slight separation of charges gives the entire molecule an electrical polarity, so water molecules are dipolar.</span>
4 0
3 years ago
A particular laser consumes 130.0 Watts of electrical power and produces a stream of 2.67×1019 1017 nm photons per second.
solniwko [45]

The missing question is:

<em>What is the percent efficiency of the laser in converting electrical power to light?</em>

The percent efficiency of the laser that consumes 130.0 Watt of electrical power and produces a stream of 2.67 × 10¹⁹ 1017 nm photons per second, is 1.34%.

A particular laser consumes 130.0 Watt (P) of electrical power. The energy input (Ei) in 1 second (t) is:

Ei = P \times t = 130.0 J/s \times 1 s = 130.0 J

The laser produced photons with a wavelength (λ) of 1017 nm. We can calculate the energy (E) of each photon using the Planck-Einstein's relation.

E = \frac{h \times c }{\lambda }

where,

  • h: Planck's constant
  • c: speed of light

E = \frac{h \times c }{\lambda } = \frac{6.63 \times 10^{-34}J.s  \times 3.00 \times 10^{8} m/s }{1017 \times 10^{-9} m }= 6.52 \times 10^{-20} J

The energy of 1 photon is 6.52 × 10⁻²⁰ J. The energy of 2.67 × 10¹⁹ photons (Energy output = Eo) is:

\frac{6.52 \times 10^{-20} J}{photon} \times 2.67 \times 10^{19} photon = 1.74 J

The percent efficiency of the laser is the ratio of the energy output to the energy input, times 100.

Ef = \frac{Eo}{Ei} \times 100\% = \frac{1.74J}{130.0J} \times 100\% = 1.34\%

The percent efficiency of the laser that consumes 130.0 Watt of electrical power and produces a stream of 2.67 × 10¹⁹ 1017 nm photons per second, is 1.34%.

You can learn more about lasers here: brainly.com/question/4869798

8 0
3 years ago
both sides in the war used military observers in hydrogen-filled balloons to watch the enemy's movements
lesya692 [45]

Answer:

Yes.

Explanation:

Hydrogen-filled balloons were widely used by the militaries during World War I (1914–1918). The main purpose of these hydrogen-filled balloons to detect movements of enemy troops and to provide direction to the artillery fire. Balloons were the targets of opposing aircraft because they knew the purpose of these balloons so they hit it whenever seen by the enemies so we can say that both sides used hydrogen-filled balloons as military observer to watch the enemy's movements.

7 0
3 years ago
How does gas pressure differ from vapor pressure
Natasha2012 [34]
The main difference between gas pressure and vapour pressure is that gas pressure is exerted by the gases above the surface of a substance whereas vapour pressure is exerted by liquids above the surface of a substance
3 0
4 years ago
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