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soldi70 [24.7K]

2 years ago

13

Look again at the pictures of the Marina District and Candlestick Park. They showed the very different effects of the same earth

quake on two locations. What connection might there be between the type of soil and the effect an earthquake has on buildings in the area?

1 answer:

lyudmila [28]2 years ago

5 0

Explanation:

The deeper the sediment layer above bedrock, the more soft soil there is for the seismic waves to travel through. Soft soil means bigger waves and stronger amplification. The earthquake damage to this building may have been influenced by the type of soil it's sitting on.

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Analyze the image below and answer the question that follows.

goldenfox [79]

There are many principles that are classified under the Gestalt Principles. The Gestalt principles believe that any stimulus can be viewed in a very simple form. According to the given image, I can say that the square represents the Gestalt principle of PROXIMITY. Hope this helps.

7 0

3 years ago

Say that you are in a large room at temperature TC = 300 K. Someone gives you a pot of hot soup at a temperature of TH = 340 K.

DiKsa [7]

Answer:0.061

Explanation:

Given

$T_C=300 k$

Temperature of soup $T_H=340 K$

heat capacity of soup $c_v=33 J/K$

Here Temperature of soup is constantly decreasing

suppose T is the temperature of soup at any instant

efficiency is given by

$\eta =\frac{dW}{Q}=1-\frac{T_C}{T}$

$dW=Q(1-\frac{T_C}{T})$

$dW=c_v(1-\frac{T_C}{T})dT$

integrating From $T_H$ to $T_C$

$\int dW=\int_{T_C}^{T_H}c_v(1-\frac{T_C}{T})dT$

$W=\int_{T_C}^{T_H}33\cdot (1-\frac{300}{T})dT$

$W=c_v\left [ T-T_C\ln T\right ]_{T_H}^{T_C}$

$W=c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]$

Now heat lost by soup is given by

$Q=c_v(T_C-T_H)$

Fraction of the total heat that is lost by the soup can be turned is given by

$=\frac{W}{Q}$

$=\frac{c_v\left [ \left ( T_C-T_H\right )-T_C\left ( \ln \frac{T_C}{T_H}\right )\right ]}{c_v(T_C-T_H)}$

$=\frac{T_C-T_H-T_C\ln (\frac{T_C}{T_H})}{T_C-T_H}$

$=\frac{300-340-300\ln (\frac{300}{340})}{300-340}$

$=\frac{-40+37.548}{-40}$

$=0.061$

4 0

3 years ago

a bends each wavelength of white light slightly differently so that each wavelength color comes out separated forming a rainbow

lianna [129]

The answer here is prism. The light passing through prism experiences bending of its multiple wavelength composition which allows it to visibly shows the difference in each of the light's color wavelength, violet bending the most while the least is the color red.

3 0

2 years ago

You just calibrated a constant volume gas thermometer. The pressure of the gas inside the thermometer is 286.0 kPa when the ther

diamong [38]

Answer:

$T_{2} = 606.69 K$

Explanation:

In that the gas thermometer is a constant volume, it is satisfied that:

$\frac{P_{1} }{T_{1} } = \frac{P_{2} }{T_{2} }$

How the boiling water is under regular atmospheric pressure, then

$T_{1} = 373 .15 K$

Thus

$\frac{286000}{373.15} = \frac{465000}{T_{2} }$

$T_{2} = 606.69 K$

5 0

2 years ago

two barges full of salted toad guts have a collision. the red barge has a mass of 150000kg and is traveling northwest at 0.25m/s

solniwko [45]

The final velocity of the red barge in the collision elastic is 0.311 m/s when it collides with blue barge pf mass 1000000 kg.

Final velocity(v3) of the red barge is calculated by following formula

m1×v1+ m2×v2= (m1+m2)v3

Substituting the value of m1= 150000 kg, v1= 0.25 m/s, m2= 1000000 kg, v2= 0.32 m/s

150000 × 0.25+ 1000000×0.32= (150000+1000000)×v3

37500+ 320000= 1150000×v3

357500= 1150000×v3

v3= 0.311 m/s

<h3>What is elastic collision velocity? </h3>

The velocity of the target particle after a head-on elastic impact in which the projectile is significantly more massive than the target will be roughly double that of the projectile, but the projectile velocity will remain virtually unaltered.

For more information on elastic collision velocity kindly visit to

brainly.com/question/29051562

#SPJ9

6 0

1 year ago