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arsen [322]
3 years ago
7

What are the three main types of stress in rock

Physics
1 answer:
Rzqust [24]3 years ago
7 0
Compression, shearing and tension are the three main types of stresses faced by rocks. In regards to the force of compression, external force acts on the rock to squeeze it until it disintegrates. In case of shearing, the rock is pulled in two opposite directions by force. This ultimately results in the rock breaking down into pieces. Tension is another force that pulls the rock, making it thin in the middle. This way the rock disintegrates as it becomes weaker and weaker in the middle. Most of the rocks present in the earths crust are hugely affected by these three forces.
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Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum
cestrela7 [59]

Answer:

T_{2}=278.80 K

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}.

Now, let's use the ideal gas equation to the initial and the final state:

\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}

Let's recall that the term nR is a constant. That is why we can match these equations.  

We can find a relation between the volumes of the initial and the final state.

\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}

Combining this equation with the first equation we have:

(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}

(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}

Now, we just need to solve this equation for T₂.

T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40

Finally, T2 will be:

T_{2}=278.80 K

6 0
3 years ago
A ball is projected upward at time t=0.0s, from a point on a roof 90m above the ground. The ball rises, then falls and strikes t
chubhunter [2.5K]
As v becomes zero at the highest point, i prefer considering different travelling directions so it will become less complicated.
dont forget to add the total time up .

also to master the skills, write down the "uvsat" may help (thats the way i found it easier to handle problems)

4 0
3 years ago
Rebecca heated 50mL of water from 0 degrees Celsius to 60 degrees Celsius. How much energy did she use to heat the water? Rememb
GuDViN [60]

Answer:

Q = 12540  J

Explanation:

It is given that,

Mass of water, m = 50 mL = 50 g

It is heated from 0 degrees Celsius to 60 degrees Celsius.

We need to find the energy required to heat the water. The formula use to find it as follows :

Q=mc\Delta T

Where c is the specific heat of water, c = 4.18 J/g°C

Put all the values,

Q=50\times 4.18\times (60-0)\\Q=12540\ J

So, 12540 J of energy is used to heat the water.

7 0
3 years ago
12. A frain moves from rest to a speed of 25 m/s in 30.0 seconds. What is its acceleration?
ziro4ka [17]
  • initial velocity=u=0m/s
  • Final velocity=v=25m/s
  • Time=t=30s

\\ \tt\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \tt\longmapsto Acceleration=\dfrac{25-0}{30}

\\ \tt\longmapsto Acceleration=\dfrac{25}{30}

\\ \tt\longmapsto Acceleration=0.8m/s^2

6 0
3 years ago
As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p
coldgirl [10]

Answer:

Explanation:

charge, q = 1.6 x 10^-19 C

distance, r = 911 nm = 911 x 10^-9 m

The Coulomb's force is given by

F=\frac{Kq_{1}q_{2}}{r^{2}}

F=\frac{9\times 10^{9}\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{\left (911\times 10^{-9}  \right )^2}

F = 2.78 x 10^-16 N

The force between the electron and the proton is 2.78 x 10^-16 N.

4 0
3 years ago
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