The following value below are the answers for letters A, B, C, and D. Please ans

Question

The following value below are the answers for letters A, B, C, and D. Please answer other questions from letter F until M.
A. Observed Gravity = 9796300 g.u.
B. Latitude Correction = 9.78436623
C. Free-Air Correction = 445mGal
D. Free-Air Anomaly =1639 mGal

13) You are standing at Burnham Park, elevation 1.442m, fora fun-filled (ha ha) gravimetric field exercise. The park is at N16 3sor and longitude E120 1". Your gravimeter gave an initial reading of 9.79630 m/s a) what is the value of your observed gravity reading in gals, mgals and gun.'s? use gua for the ensuing questions in this problem. pts) b) what is the value of gravity (go) the International formula predicts at your location? (2 pts) c) What is the value of the free air correction at your site? pts) d Calculate the free air gravity anomaly. (2 pts) e) Assuming that the underlying mass has an a (2 pts) f) Calculate the Bouguer anomaly. (2 pts) according to Airv s m io Assuming that the mass you are surveying as in perfect g/cc? (2 pts) mantle has a density of out how thick a the find yourself at Burnham Park, but found root should this you height. What would h) suppose that, time- raven irv: to the distant future. ers from its previ balance. B pts that the place has already been eroded, removing 10a to restore isostatic obtain happen isostatically? the am vaunt of you are supposed to You came back to the present after the bad a and remembered that readiness the of the You dream followicag graph alonit center the 50, 100 50 What does the positive Bougue reveal about the density of the ore body? (1 pty k) What can you say about the target? Give 2 possible interpretations and quantify the depth pts) Illustrate your What extra gravity data would be needed to discriminate between the two possibilities? answer m., th m) Assume that the density contrast between the orebody and the host rock is sookercu estimate radius of the deposit. (4 pts)

Explanation / Answer

Answer- Bouguer gravity anomoly = obserbed gravity - latitude correction + free air correction - Terrain correction

= 9796300 - 9.78436623+ 0.445- 1.639

=9696289.0 g.u

answer- j- the positive bouguer gravity anomoly reveal the density about the object. the greater will be value of positive bouguer gravity anomoly higher will be density of the object or rock in subsurface. higher gravity anomoly represent denser rock.

answer- k- some possible interpretation about the target by the value of bouguer gravity anomoly are following

- target is denser rock representing higher value of bouguer gravity anomoly than surrounding. the surrounding rock are less denser representing low value of bouguer gravity anomoly.

-Bouguer anomaly impression of subsurface density. Low (negative) values of Bouguer anomaly indicate lower density beneath the measurement point. High (positive) values of Bouguer anomaly indicate higher density beneath the measurement point

- gravity method is not at depth of basement rock estimation

answer- the extra data that is required for discriminate between target and background value are following

- free air gravity value

- latitude correction value

- elevation or altitude or terrain correction value

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