Question 2 help me and i'll give u brainlest

Describe the electron transfers that occur in the formation of calcium fluoride from elemental calcium and elemental fluorine.

boyakko [2]

Answer:

Check explanation

Explanation:

In the formation of calcium fluoride we take calcium and fluorine.

in elemental form calcium exist in solid form and fluorine in gaseous form.

formation of compound takes place to complete their octet, in case of calcium need to remove two electron and need to add one elecron in fluorine to complete their octet so two electron will ransferred from calcium to two fluorine atom.

3 0

3 years ago

What is the difference between balanced force and action reaction force

Vilka [71]

Answer:

Balanced forces are equal and opposite forces that act on the same object. ... Action-reaction forces are equal and opposite forces that act on different objects, so they don't cancel out. In fact, they often result in motion.

5 0

3 years ago

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True or false gravitational pull decreases with an increase of distance between two objects

ohaa [14]

Answer:

true! : )

(i underlined the place where the answer is the other information is just as important but if you do not want to read it you do not have to)

Explanation:

Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases. the greater the mass, the greater the gravitational pull. gravitational pull decreases with an increase in the distance between two objects. Since gravitational force is inversely proportional to the square of the separation distance between the two interacting objects, more separation distance will result in weaker gravitational forces. So as two objects are separated from each other, the force of gravitational attraction between them also decreases.

5 0

3 years ago

When running a half marathon (13.1 miles), it took Kevin 8 minutes to run from mile marker 1 to mile marker 2, and 18 minutes to

Vsevolod [243]

Answer:

It took Kevin 26 minutes to run from markers 1 to 4

His average speed from mile markers 1 to 4 is 0.154 miles/minute

Kevin must run by average speed 0.1 miles/minute to finish the race

Explanation:

Lets explain how to solve the problem

A half marathon 13.1 miles

Kevin took 8 minutes to run from mile marker 1 to mile marker 2 and

18 minutes to run from mile marker 2 to mile marker 4

→ He took 8 minutes and 18 minutes to run from marker 1 to marker 4

→ The total time of the first 4 marker = 8 + 18 = 26 minutes

It took Kevin 26 minutes to run from markers 1 to 4

Average speed is total distance divided by total time

The average speed of Kevin as he ran from mile marker 1 to mile

marker 4 is the 4 miles divides by 26 minutes

→ Average speed = 4 ÷ 26 = $\frac{2}{13}$ = 0.154 miles/minute

His average speed from mile markers 1 to 4 is 0.154 miles/minute

It took Kevin 71 minutes to pass mile marker 9

Kevin need to complete the race in 112 minutes, then what must

Kevin's average speed be as he travels from mile marker 9 to the

finish line?

The total distance of the race is 13.1 miles, he ran 9 miles

→ The remaining distance = 13.1 - 9 = 4.1 miles

He must run 4.1 miles to complete the race

The total time is 112 minutes, he used 71 minutes to run the first 9 miles

→ The remaining time = 112 - 71 = 41 minutes

He must finish the 4.1 miles in 41 minutes

→ His average speed for the last part of the race = 4.1 ÷ 41 = 0.1 mi/min

Kevin must run by average speed 0.1 miles/minute to finish the race

7 0

3 years ago

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A boat radioed a distress call to a Coast Guard station. At the time of the call, a vector A from the station to the boat had a

VashaNatasha [74]

Answer:

d = 39.7 km

Explanation:

initial position of the boat is 45 km away at an angle of 15 degree East of North

so we will have

$r_1 = 45 sin15 \hat i + 45 cos15 \hat j$

$r_1 = 11.64 \hat i + 43.46\hat j$

after some time the final position of the boat is found at 30 km at 15 Degree North of East

so we have

$r_2 = 30 cos15\hat i + 30 sin15 \hat j$

$r_2 = 28.98\hat i + 7.76 \hat j$

now the displacement of the boat is given as

$d = r_2 - r_1$

$d = (28.98\hat i + 7.76 \hat j) - (11.64 \hat i + 43.46\hat j)$

$d = 17.34 \hat i - 35.7 \hat j$

so the magnitude is given as

$d = \sqrt{17.34^2 + 35.7^2}$

$d = 39.7 km$

4 0

3 years ago