6x10^8 kg<br> —————<br> 2x10^4 m^3

KIM [24]

The appropriate answer is c. geysers. A geysers a hot water fountain that spouts intermittently with great force, frequently accompanied by a thunderous roar. The world famous Old Faithful is located in Yellowstone National Park. This geyser erupts every 65 min sending a jet of water almost 60 meters into the air. Sinkholes and caves are formed by the action of groundwater on carbonate rocks which causes them to colapse and former these structures.

6 0

3 years ago

Read 2 more answers

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

2 years ago

Which of the following changes would be a physical change to a substance?(1 point)

Naddika [18.5K]

Answer:

all of them I think

Explanation:

6 0

2 years ago

Why should you do you should

uranmaximum [27]

One does not simply should you do you should

8 0

2 years ago

Find the gravitational potential energy of an 84 kg person standing atop Mt. Everest at an altitude of 8848 m. Use sea level as

djverab [1.8K]

Answer:

$E=7.28\times 10^6\ J$

Explanation:

Given that,

Mass of a person, m = 84 kg

The person is standing at a top of Mt. Everest at an altitude of 8848 m

We need to find the gravitational potential energy of the person. We know that the gravitational potential energy is possessed due to the position of an object. It is given by :

E = mgh, g is the acceleration due to gravity

$E=84\ kg\times 9.8\ m/s^2\times 8848\ m\\\\E=7283673.6\ J\\\\E=7.28\times 10^6\ J$

So, the gravitational potential energy of the person is $7.28\times 10^6\ J$

6 0

2 years ago

6x10^8 kg ————— 2x10^4 m^3