Help me!!! Due todayyyyy!!!!!!!!!!!!!!! ❤️ :)

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

4 years ago

WILL MARK BRAINLIST

ryzh [129]

Answer:

Definition of envelop

transitive verb

1: to enclose or enfold completely with or as if with a covering

2: to mount an attack on (an enemy's flank)

3 0

2 years ago

HELP PLEASE!!! 30+ points!!

GarryVolchara [31]

1) 9.26 cm

Explanation:

The focal length of a plane mirror is virtually infinite. Considering the lens equation,

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where f is the focal length, p is the object distance, q the image distance. If we replace f with infinity, we get

$q=-p$

The magnification equation states that

$y' = -\frac{q}{p}y$

where y is the size of the object and y' the size of the image. Substituting q=-p, we get

$y'=y$

this means that the image produced by a plane mirror is always:

- Upright (y' is positive)

- The same size as the object

In this case, we have a book of height 9.26 cm (y=9.26 cm). This means that the magnitude of the size of the image (y') will be 9.26 cm as well.

2) 22.7 cm

As we said before, due to the infinite focal length of a plane mirror,

$q=-p$

this means that the image produced by a plane mirror is always:

- Virtual (because q is negative)

- At the same distance from the mirror as the object

In this case, we have a book placed at 22.7 cm from the mirror (p=22.7 cm). This means that the magnitude of the distance of the image from the mirror (q) will be 22.7 cm as well.

3) 1.60 m/s

We said previously that the image produced by a plane mirror is always at the same distance from the mirror as the real object. This implies that whenever we move the object toward/away from the mirror, the distance p will alway remain equal to the distance q. But this also means that the object and the distance are moving toward/away from the mirror at the same speed.

Therefore, since in this case the person is moving away from the mirror at 1.60 m/s, the image will also move away at a speed of 1.60 m/s.

4 0

3 years ago

HELP!!!!!! Need Help !!!

denis23 [38]

Answer:

its y

Explanation:

3 0

3 years ago

A train at a constant 60.0 km/h moves east for 40.0 min, then in a direction 50.0 degrees east of due north for 20.0 min, and th

son4ous [18]