PLEASE HELP 25 POINTS (WILL AWARD BRAINIEST ANSWER)

At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107

dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

$K = - V\cdot \frac{dP}{dV}$

Where:

$K$ - Bulk module, measured in pascals.

$V$ - Sample volume, measured in cubic meters.

$P$ - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

$-\frac{K \,dV}{V} = dP$

This resultant expression is solved by definite integration and algebraic handling:

$-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP$

$-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}$

$\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}$

$\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }$

The final volume is predicted by:

$V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }$

If $V_{o} = 1\,m^{3}$ , $P_{o} - P_{f} = -10132500\,Pa$ and $K = 2.3\times 10^{9}\,Pa$ , then:

$V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }$

$V_{f} \approx 0.996\,m^{3}$

Change in volume due to increasure on pressure is:

$\Delta V = V_{o} - V_{f}$

$\Delta V = 1\,m^{3} - 0.996\,m^{3}$

$\Delta V = 0.004\,m^{3}$

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

8 0

3 years ago

What is the connection between angle of incidence and angle of reflection

Radda [10]

Answer:

The normal line divides the angle between the incident ray and the reflected ray into two equal angles. The angle between the incident ray and the normal is known as the angle of incidence. The angle between the reflected ray and the normal is known as the angle of reflection.

8 0

3 years ago

A ball is thrown at 20 m/s from the ground upwards at an angle of elevation of 30°. How far away does it land? 35.35 m

mestny [16]

Answer:

35.35 m

Explanation:

The following data were obtained from the question:

Initial velocity (u) = 20 m/s

Angle of projection (θ) = 30°

Acceleration due to gravity (g) = 9.8 m/s²

Range (R) =.?

The range (i.e how far away) of the ball can be obtained as follow:

R = u² Sine 2θ /g

R = 20² Sine (2×30) / 9.8

R = 400 Sine 60 / 9.8

R = (400 × 0866) / 9.8

R = 346.4 / 9.8

R = 35.35 m

Therefore, the range (i.e how far away) of the ball is 35.35 m

8 0

3 years ago

What did solomon asch discover in his famous experiment on judging the lengths of lines?

BaLLatris [955]

Answer:

Asch (1956) found that group size influenced whether subjects conformed. The bigger the majority group (no of confederates), the more people conformed, but only up to a certain point.

Explanation:

4 0

3 years ago

Air pressure increases as you travel higher above sea level. This is the reason that cabins in commercial airliners require pres

irakobra [83]

The answer is true about the cabins in commercial airliners that require pressurization.

<h3>Why are the cabins of commercial airplanes pressurized?</h3>

Airplanes are pressurized because the air is very thin at the high altitude where they fly. The passenger jet has a cruising altitude of about 30,000 - 40,000 feet. At this altitude or height, humans can't breathe very well and our body gets less amount of oxygen. Most aircraft cabins are pressurized to an altitude about 8,000 feet. This is called cabin altitude. Aircraft pilots have access to the control's mode of a cabin pressure control system and if needed it can command the cabin to depressurize.

So we can conclude that cabins in commercial airliners require pressurization because of the greater pressure of the surrounding environment.

Learn more about pressure here: brainly.com/question/28012687

#SPJ1

4 0

1 year ago