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2 years ago

Compared with space operations specialists, intelligence officers are which of the following?

Engineering

1 answer:

puteri [66]2 years ago

7 0

Answer:

whats the answer if you did it?

Explanation:

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function summedValue = SummationWithLoop(userNum) % Summation of all values from 1 to userNum summedValue = 0; i = 1; % Write a

Alexeev081 [22]

Answer:

function summedValue = SummationWithLoop(userNum)

% Summation of all values from 1 to userNum

summedValue = 0;

i = 0;

% use a while loop that assigns summedValue with the

% sum of all values from 1 to userNum

while(i <= userNum)

summedValue = summedValue + i;

i = i + 1;

end

8 0

3 years ago

Are the wooden pillars shown in the image below, a dead load?

Brrunno [24]

Answer:

Explanation:

it's not a dead load because when load is put on the pillars it's not fully straining it's been slowly getting to be heavier in that period of time before it falls

6 0

3 years ago

Solid Isomorphous alloys strength

Sphinxa [80]

Answer:

Explanation:

ℎ

3 0

3 years ago

A 11.5 nC charge is at x = 0cm and a -1.2 nC charge is at x = 3 cm ..At what position or positions on the x-axis is the electric

diamong [38]

Answer:

Explanation:

Given

$q_1=11.5\ nC$ charge is placed at $x=0\ cm$

another charge of $q_2=-1.2\ nC$ is at $x=3\ cm$

We know that Electric field due to positive charge is away from it and Electric field due to negative charge is towards it.

so net electric field is zero somewhere beyond negatively charged particle

Electric Field due to $q_2$ at some distance r from it

$E_2=\frac{kq_2}{r^2}$

Now Electric Field due to $q_1$ is

$E_1=\frac{kq_1}{(3+r)^2}$

Now $E_1+E_2=0$

$\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0$

$\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}$

$\frac{3+r}{r}=3.095$

thus $r=1.43\ cm$

Thus Electric field is zero at some distance r=1.43 cm right of $q_2$

3 0

3 years ago

When would working with machinery be a common type of caught-in and caught-between<br> hazard?

tigry1 [53]

Answer:

A working with machinery be a common type of caught-in and caught-between hazard is described below in complete detail.

Explanation:

“Caught in-between” accidents kill mechanics in a variety of techniques. These incorporate cave-ins and other hazards of tunneling activity; body parts extracted into unconscious machinery; reaching within the swing range of cranes and other installation material; caught between machine & fixed objects.

6 0

2 years ago