The question is in the image

You get to the dive site, put on your gear, and are ready to go under. You check your scuba tank. It is filled to a pressure of

Dennis_Churaev [7]

Answer:

If you were to increase the volume the tank would get heavy.

Explanation:

8 0

3 years ago

A cell membrane has a thickness of about 7 nm. How many cell membranes would it take to make a stack 3.4 in high?

sergiy2304 [10]

Answer:

486 of cell membranes

Explanation:

one cell membrane has 7*10^-9m as the height

So, by dividing 3.4 with 7*10^-9 you will get the number of cell require to form the height.

3.4/(7*10^-9)=485.71

because we can't have decimal number of cell we would round it into three significant figures, 486.

3 0

3 years ago

A parent swings a 13.5 kg child

solmaris [256]

The speed of the child is 2.21 m/s

Explanation:

The child is moving in circular motion. First of all, we find its frequency of revolution, which is given by:

$f=\frac{N}{t}$

where

N = 11 is the number of revolutions

t = 25.4 s is the time elapsed

So,

$f=\frac{11}{25.4}=0.433 Hz$

Now we find the angular frequency, which is given by

$\omega = 2\pi f = 2\pi(0.433)=2.72 rad/s$

And now we can fidn the linear speed of the child, which is given by

$v=\omega r$

where

$\omega=2.72 rad/s$ is the angular frequency

r = 0.811 m is the radius of the circular trajectory

Substituting,

$v=(2.72)(0.811)=2.21 m/s$

Learn more about circular motion:

brainly.com/question/2562955

brainly.com/question/6372960

#LearnwithBrainly

3 0

3 years ago

I don't understand any of this please help !

OlgaM077 [116]

You can see what is the electron configuration by looking at the layout of the periodic tables. the first shell will have a max of 2 electrons on it, once the first one is filled up a second is added with a max of 8 electrons on it and so on with the 8 as a max. so He, and H will only have them on the first shell but every horizontal row is a new valence or outer shell. so lets say for carbon look at the number in the upper left corner of the box will tell you the total number of electrons you will need. so start off with the first two electrons on the first shell. now you know that carbon needs 6 electrons in total, since you can only have a max of 2 on the first shell you need a second one so on the second one you will have to have the remaining 4. now elements are most stable when they have a full valence shell becuase those are the only electrons that will react with others. so if carbon has 4 it wants to either gain or lose 4 electrons so you could say that it would bond with 4H since each H will donate 1 electron to the C valence shell making all the H and C stable. CH4(methane)

6 0

3 years ago

coin 1 is thrown upward from the top of 100m tower with a speed of 15m/s. coin 2 is dropped from the top of the tower 2.0second

kvv77 [185]

The height below the tower at which coin 1 pass coin 2 is 89.04 m.

The given parameters:

height of the tower, h = 100 m

initial velocity of coin 1, v = 15 m/s

time spent in air by coin 1 before coin 2 was dropped = 2s

To find:

the height below the tower at which coin 1 passes coin 2

Find the maximum height attained by coin 1 before falling to the ground:

$v^2 = u^2 - 2gh\\\\where;\\\\v \ is \ the \ final \ velocity \ of \ coin \ 1 \ at \ maximum \ height, v \ = 0\\\\0 = (15^2) - 2(10)h\\\\20h = 225\\\\h = \frac{225}{20} \\\\h = 11.25 \ m$

Find the time taken for coin 1 to fall to the ground:

Total height of coin 1 above the ground, H = 11.25 m + 100 m = 111.25 m

$t = \sqrt{\frac{2H}{g} } \\\\t = \sqrt{\frac{2\times 111.25}{10} } \\\\t = 4.72 \ s$

But the time taken for the coin 1 to reach 11.25 m above the tower:

$t_1 = \sqrt{\frac{2h}{g} } \\\\t_1 = \sqrt{\frac{2\times 11.25}{10} } \\\\t_1 = 1.5 \ s$

Total time spent by coin 1 before reaching ground with respect to coin 2:

time = (1.5 s + 4.72 s) - 2 s

time = 4.22 s

<u>Note:</u><em> the 2 s was subtracted to keep both coins at a fair starting time below the tower</em>.

Find the total time taken for coin 2 to fall to the ground:

Height of coin 2 above the ground = 100 m

Total time taken by coin 2 before falling to the ground is calculated as:

$t_2 = \sqrt{\frac{2(100)}{10} } \\\\t_2 = 4.47s$

The time at which coin 1 will pass coin 2 is 4.22 s.

Find the height below the tower when the time is 4.22 s.

$h = \frac{1}{2} (10)(4.22)^2\\\\h = 89.04 \ m$

Thus, the height below the tower at which coin 1 pass coin 2 is 89.04 m.

Learn more here: brainly.com/question/18943273

6 0

3 years ago