Mudflows composed of soil, volcanic debris, and water can occur as the result of an explosive volcanic eruption. What are these mudflows called?

Answers

Answer 1

Answer:

These mudflows are called Lahar.

A quickly moving mixture of rock debris and water that begins on a volcano's slopes is referred to as a lahar in Indonesian. Other names for lahars are volcanic mudflows and debris floods. The size, pace, and volume of material transported by a moving lahar can constantly alter as it rushes downstream. It resembles a swirling slurry of wet concrete.

The melting of snow and ice as well as the ingestion of river or lake water by the moving slurry may both add to its water consumption. A lahar's starting flow may be quite tiny, but as it entrains and integrates everything in its path, including rocks, dirt, vegetation, even structures like houses and bridges, it may increase in volume.

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A large aquarium has portholes of thin transparent plastic with a radius of curvature of 1.95 m and their convex sides facing into the water. A shark hovers in front of a porthole, sizing up the dinner prospects outside the tank.a) If one of the sharks teeth is exactly 46.5 cm from the plastic, how far from the plastic does it appear to be to observers outside the tank? (You can ignore refraction due to the plastic.)b) Does the shark appear to be right side up or upside down?c) If the tooth has an actual length of 5.00 cm, how long does it appear to the observers?

Answers

Answer:

Explanation:

For refraction through a curved surface , the formula is as follows

μ₂ / v - μ₁ / u = (μ₂ -μ₁ )/R , Here μ₂( air) = 1 , μ₁ ( water) = 4/3 , R = 1.95 m

u , object distance = - .465 m

1 / v + 1.333 / .465 = (1 -1.333 )/1.95

1 / v + 2.8667 = - .171

1 / v = - 2.8667 - .171 = - 3.0377

v = - .3292 m

= - 32.92 cm

image will be formed in water.

c ) magnification = μ₁v / μ₂u , μ₁ = 1.33 , μ₂ = 1 , u = 46.5 , v = 32.92 .

= (1.33 x 32.92) / (1 x 46.5)

= .94

size of image of teeth = .94 x 5

= 4.7 cm .

We had a homework problem in which the Arrhenius equation was applied to the blinking of fireflies. Several other natural phenomena also obey that equation, including the temperature dependent chirping of crickets. A particular species, the snowy tree cricket, has been widely studied. These crickets chirp at a rate of 178 times per minute at 25.0°C, and the activation energy for the chirping process is 53.9 kJ/mol. What is the temperature if the crickets chirp at a rate of 126 times per minute?

Answers

Answer:

Temperature = 20.35°C

Explanation:

Arrhenius equation is as follows:

k = A*exp(-Ea/(R*T)), where

k = rate of chirps

Ea = Activation Energy

R = Universal Gas Constant

T = Temperature (in Kelvin)

A = Constant

Given Data

Ea = 53.9*10^3 J/mol

R = 8.3145 J/(mol.K)

T = 273.15 + 25 K

k = 178 chirps per minutes

Calculation

Using the Arrhenius equation, we can find A,

A= 4.935x10^11

Now we can apply the same equation with the data below to find T at k=126,

k = A*exp(-Ea/(R*T))

Ea = 53.9*10^3

R = 8.3145

k = 126

T = 20.35°C

A power P is required to do work W in a time interval T. What power is required to do work 3W in a time interval 5T? (a) 3P (b) 5P (c) 3P/5 (a) P (e) 5P/3

Answers

Answer:

Explanation:

The formula to calculate the power is:

$P=(W)/(T)$

where

W is the work done

T is the time required for the work to be done

In the second part of the problem, we have

Work done: 3W

Time interval: 5T

So the power required is

$P=(3W)/(5T)=(3)/(5)(W)/(T)=(3)/(5)P$

Air enters an adiabatic compressor at 104 kPa and 292 K and exits at a temperature of 565 K. Determine the power (kW) for the compressor if the inlet volumetric flow rate is 0.15 m3/s. Use constant specific heats evaluated at 300 K.

Answers

Answer:

$\dot W_(in) = 49.386\,kW$

Explanation:

An adiabatic compressor is modelled as follows by using the First Law of Thermodynamics:

$\dot W_(in) + \dot m \cdot c_(p)\cdot (T_(1)-T_(2)) = 0$

The power consumed by the compressor can be calculated by the following expression:

$\dot W_(in) = \dot m \cdot c_(v)\cdot (T_(2)-T_(1))$

Let consider that air behaves ideally. The density of air at inlet is:

$P\cdot V = n\cdot R_(u)\cdot T$

$P\cdot V = (m)/(M)\cdot R_(u)\cdot T$

$\rho = (P\cdot M)/(R_(u)\cdot T)$

$\rho = ((104\,kPa)\cdot (28.02\,(kg)/(kmol)))/((8.315\,(kPa\cdot m^(3))/(kmol\cdot K) )\cdot (292\,K))$

$\rho = 1.2\,(kg)/(m^(3))$

The mass flow through compressor is:

$\dot m = \rho \cdot \dot V$

$\dot m = (1.2\,(kg)/(m^(3)))\cdot (0.15\,(m^(3))/(s) )$

$\dot m = 0.18\,(kg)/(s)$

The work input is:

$\dot W_(in) = (0.18\,(kg)/(s) )\cdot (1.005\,(kJ)/(kg\cdot K))\cdot (565\,K-292\,K)$

$\dot W_(in) = 49.386\,kW$

Consider the following statements. A. Heat flows from an object at higher temperature to an object at lower temperature; B. Heat flows from an object in liquid state to an object in solid state; C. Heat flows from an object with higher thermal energy to one with lower thermal energy. Which statements are true? 1. A only 2. A and C only 3. B and C only 4. None is true.

Answers

Only ' A ' is always true. (choice-1)

' B ' is not true when you drop a red hot spoon into cold soup.

' C ' is not true when you drop a red hot marble into a cool swimming pool.

How many hours does earth take to complete one rotation?

Answers

Answer:

24 hours take earth to complete rotation

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