What type of data allows geologists to know what rock layers lie underneath Minnesota?

1 answer:

5 0

The answer is well log data, it is a detailed log of information taken from a borehole which geologist used to study geological formations of the earth's layer taken from samples returned from the borehole which was dugged.

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What structures can be found in the axial region of the body?

Softa [21]

Answer:

The axial region of the body consists of the bones of the head, trunk of a vertebrate, skull, vertebral column, and thoracic cage. The human skeleton consists of 80 bones.

Explanation:

The axial region of the body consists of the bones of the head, trunk of a vertebrate, skull, vertebral column, and thoracic cage. The human skeleton consists of 80 bones.

It is composed of the following six parts:

1. Skull (22 bones)

2. Ossicles of the middle ear

3. Hyoid bone

4. Rib cage

5. Sternum

6. Vertebral column

The axial region of the body forms the vertical axis of the body as the axial skeleton supports the head, neck, back, and chest.

3 0

3 years ago

The vapor pressure of benzene, C6H6, is 40.1 mmHg at 7.6°C. What is its vapor pressure at 60.6°C? The molar heat of vaporization

ANEK [815]

Answer:

The vapor pressure at 60.6°C is 330.89 mmHg

Explanation:

Applying Clausius Clapeyron Equation

$ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]$

Where;

P₂ is the final vapor pressure of benzene = ?

P₁ is the initial vapor pressure of benzene = 40.1 mmHg

T₂ is the final temperature of benzene = 60.6°C = 333.6 K

T₁ is the initial temperature of benzene = 7.6°C = 280.6 K

ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol

R is gas rate = 8.314 J/mol.k

$ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg$

Therefore, the vapor pressure at 60.6°C is 330.89 mmHg

6 0

3 years ago

Read 2 more answers

During a very quick stop, a car decelerates at 7.6 m/s2. Assume the forward motion of the car corresponds to a positive directio

Mashcka [7]

Answer:

24.57 revolutions

Explanation:

(a) If they do not slip on the pavement, then the angular acceleration is

$\alpha = a / r = 7.6 / 0.26 = 29.23 rad/s^2$

(b) We can use the following equation of motion to find out the angle traveled by the wheel before coming to rest:

$\omega^2 - \omega_0^2 = 2\alpha\Delta \theta$

where v = 0 m/s is the final angular velocity of the wheel when it stops, $\omega_0$ = 95rad/s is the initial angular velocity of the wheel, $\alpha = -29.23 rad/s^2$ is the deceleration of the wheel, and $\Delta \theta$ is the angle swept in rad, which we care looking for: