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3 years ago

"geophysical exploration definition"

Engineering

2 answers:

Strike441 [17]3 years ago

7 0

Answer:

Exploration geophysics is an applied branch of geophysics and economic geology, which uses physical methods, such as seismic, gravitational, magnetic, electrical and electromagnetic at the surface of the Earth to measure the physical properties of the subsurface, along with the anomalies in those properties

lara31 [8.8K]3 years ago

7 0

Answer:

Exploration geophysics

Explanation:

Please mark me as brainliest

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An aluminum part will be subjected to cyclic loading where the maximum stress will be 300 MPa and the minimum stress will be-100

Dominik [7]

Answer:

a) The mean stress experimented by the aluminium part is 100 megapascals, b) The stress amplitude of the aluminium part is 400 megapascals, c) The stress ratio of the aluminium part is 4.

Explanation:

a) The mean stress is determined by this expression:

$\sigma_{m} = \frac{\sigma_{min}+\sigma_{max}}{2}$

Where:

$\sigma_{m}$ - Mean stress, measured in megapascals.

$\sigma_{min}$ - Minimum stress, measured in megapascals.

$\sigma_{max}$ - Maximum stress, measured in megapascals.

If we know that $\sigma_{min} = -100\,MPa$ and $\sigma_{max} = 300\,MPa$ , the mean stress is:

$\sigma_{m} = \frac{-100\,MPa+300\,MPa}{2}$

$\sigma_{m} = 100\,MPa$

The mean stress experimented by the aluminium part is 100 megapascals.

b) The stress amplitude is given by the following difference:

$\sigma_{a} = |\sigma_{max}-\sigma_{min}|$

Where $\sigma_{a}$ is the stress amplitude, measured in megapascals.

If we know that $\sigma_{min} = -100\,MPa$ and $\sigma_{max} = 300\,MPa$ , the stress amplitude is:

$\sigma_{a} = |300\,MPa-(-100\,MPa)|$

$\sigma_{a} = 400\,MPa$

The stress amplitude of the aluminium part is 400 megapascals.

c) The stress ratio ( $R$ ) is the ratio of the stress amplitude to mean stress. That is:

$R = \frac{\sigma_{a}}{\sigma_{m}}$

If we know that $\sigma_{m} = 100\,MPa$ and $\sigma_{a} = 400\,MPa$ , the stress ratio is:

$R = \frac{400\,MPa}{100\,MPa}$

$R = 4$

The stress ratio of the aluminium part is 4.

3 0

3 years ago

Question 2

ValentinkaMS [17]

Answer:

r h ch y TC GG ch hh h h

No se

Explanation:

No seibibk h j h uh t t t

3 0

2 years ago

Ignorance: noun

jarptica [38.1K]

Answer:

A: the condition of living without the benefit of education at some level

Explanation:

As a noun, ignorance is simply defined as the lack of knowledge about a common thing.

For example, he was ignorant about the schemes of his colleagues at the office.

Thus, from the given options, the correct answer is: "the condition of living without the benefit of education at some level".

3 0

3 years ago

For a Cu-Ni alloy containing 53 wt.% Ni and 47 wt.% Cu at 1300°C, calculate the wt.% of the alloy that is solid and wt.% of allo

Zina [86]

Answer:

Hello your question is incomplete attached below is the complete question

answer: wt.% of alloy that is solid = 61.5%

wt.% of allot that is liquid = 38.5%

Explanation:

To determine the wt.% of the alloy that is solid

= $\frac{R}{R +S } * 100$

= $\frac{53-45}{58-45} * 100$ = 61.5%

To determine the wt.% of the alloy that is liquid

= $\frac{S}{S+R} * 100\\$

= $\frac{58-53}{58-45} *100$ = 38.5%

attached below is a free hand sketch as well

7 0

3 years ago

A square steel bar has a length of 8.4 ft and a 2.1 in by 2.1 in cross section and is subjected to axial tension. The final leng

nikitadnepr [17]

Answer:

Poissons ratio = -0.3367

Explanation:

Poissons ratio = Lateral Strain / Longitudinal Strain

In this case, the longitudinal strain will be:

Strain (longitudinal) = Change in length / total length

Strain (longitudinal) = (8.40392 - 8.4) / 8.4

Strain (longitudinal) = 4.666 * 10^(-4)

While the lateral strain will be:

Strain (Lateral) = Change in length / total length

Strain (Lateral) = (2.09967 - 2.1) / 2.1

Strain (Lateral) = -1.571 * 10^(-4)

Solving the poisson equation at the top we get:

Poissons ratio = -1.571 / 4.666 <u>( 10^(-4) cancels out )</u>

Poissons ratio = -0.3367

6 0

4 years ago

"geophysical exploration definition"​

"geophysical exploration definition"