rlafferty5604

04.07.2019 • 

Physics

(a) a typical thermionic emission source used in a sem has a total current of 1.6ha. if 10 % of the electron current is included in the probe, how many electrons are incident on the specimen in 10s? using this sem to image a zone-plate made of alternating layers of c6 and w74. make the you are following assumptions- (0)the secondary electron (se) yield is 0.2% for c and 0.05% for w.(ii)the back scattered electron(bse) yield is 0.12 %, where z is the atomic number. (ii) the efficiency of detection for both electrons is 10% (y) the standard deviation, o-n, where n is the counts. (b) how many secondary electrons of c and w will be detected in 1s? (c) how many bse of c and w will be detected in 1s? (d) what will be the contrasts across a w/c boundary for secondary and back-scattered electrons? (e) is the contrast statistically significant in each case? (f) which method, se or bse, would be better to image the w/c boundary? give a brief explanation.

Ответ:

daedae11142

19.05.2020

Physics

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GPE = 388.08 Joules.

Explanation:

Given the following data;

Mass = 0.550kg

Speed = 335 m/s

Height = 72 meters

We know that acceleration due to gravity, g is equal to 9.8 m/s²

To find the gravitational potential energy;

Gravitational potential energy (GPE) is an energy possessed by an object or body due to its position above the earth.

Mathematically, gravitational potential energy is given by the formula;

Where;

G.P.E represents potential energy measured in Joules.

m represents the mass of an object.

g represents acceleration due to gravity measured in meters per seconds square.

h represents the height measured in meters.

Substituting into the formula, we have;

GPE = 388.08 Joules.

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