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AnnyKZ [126]
3 years ago
15

Although it shouldn’t have happened, on a dive I fail to watch my SPG and run out of air. If my buddy is close by, my best optio

n is to __________. Another option is to _____________, if I’m in shallow water and the surface is closer than my buddy.
A ascend using my buddy's alternate air source / make a controlled emergency swimming ascent




B ascend using my buddy's alternate air source / make an buoyant emergency ascent




C make a normal ascent / ascend using my buddy's alternate air source




D make a controlled emergency swimming ascent / make a normal ascent
Physics
1 answer:
Airida [17]3 years ago
5 0

Answer:

A) ascend using my buddy's alternate air source / make a controlled emergency swimming ascent

Explanation:

When it is found that you are out of air while under water, first of all don't panic, look for your buddy. If you are unable to do that so, then you need to make an emergency ascent. First try to make a <u>Controlled Emergency Swimming Ascent</u> (CESA). This ascent remains under control and is performed at a safe ascent rate. As you ascend the air in your lungs will expand with decreasing ambient pressure. To avoid an over pressurization injury, always exhale a continuous string of bubbles while going up.

If you are not sure you will make to the surface that leading to inhale the only option is to turn the CESA into a Buoyant Emergency Ascent. To be ready locate your weight system as you ascend. As an addition remove the weight from one of weight pockets and hold it away from your body in preparation of dropping if necessary. Dropping the weight will give you an upward buoyant force which is an <u>uncontrolled buoyant emergency ascent</u> and <u><em>should be performed only as the last option</em></u>.

So, according to this, first, always have a look at your SPG. Then, if you are out of air, look for your buddy, if not found then make CESA and the last option will be buoyant emergency ascent.

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When the substance are moved close together and they move more quickly they get compressed.
4 0
3 years ago
A piano string having a mass per unit length of 5.00 g/m is under a tension of 1350 N. Determine the speed of transverse waves i
padilas [110]

Answer:

The speed of transverse waves in this string is 519.61 m/s.

Explanation:

Given that,

Mass per unit length = 5.00 g/m

Tension = 1350 N

We need to calculate the speed of transverse waves in this string

Using formula of speed of the transverse waves

v=\sqrt{\dfrac{T}{\mu}}

Where, \mu = mass per unit length

T = tension

Put the value into the formula

v = \sqrt{\dfrac{1350}{5.00\times10^{-3}}}

v =519.61\ m/s

Hence, The speed of transverse waves in this string is 519.61 m/s.

6 0
3 years ago
Someone please help !!
masha68 [24]
I’m pretty sure it’s c.... hope it helps and hope it’s right.
7 0
3 years ago
1-A car moves toward east 12km is represented as A and it turns towards south 16km is represented as B. What is the resultant ve
hjlf

1. A-20 km south east

The car's displacement consists of two components into two different directions. Using a system of coordinates in which x represents the east direction and y represents the south direction, the two displacements are:

d_x = 12 km east

d_y = 16 km south

Since the two components are orthogonal to each other, we can find the resultant displacement by using Pythagorean's theorem:

d=\sqrt{d_x^2+d_y^2}=\sqrt{(12 km)^2+(16 km)^2}=\sqrt{400}=20 km

and the direction is between the two original directions, so south-east.

2. D. 10 m/s

First of all, we need to calculate the total time the stone took to hit the ground. Since the vertical distance covered is S = 78.4 m, and since the motion is an accelerated motion with constant acceleration g=9.8 m/s^2, we have

S=\frac{1}{2}gt^2

From which we find the total time of the fall, t:

t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(78.4 m)}{9.8 m/s^2}}=4 s

Now we can consider the horizontal motion of the stone: we know that the stone travels for d = 40 m in a time of t = 4 s, therefore the horizontal velocity of the stone is

v=\frac{d}{t}=\frac{40 m}{4 s}=10 m/s

3. B=32.32 m

As in the previous problem, we have to calculate the total time it takes for the stone to reach the river first. Since the vertical distance covered is S = 20 m, we have

t=\sqrt{\frac{2S}{g}}=\sqrt{\frac{2(20 m)}{9.8 m/s^2}}=2.0 s

And since the stone is traveling horizontally at v = 16 m/s, the horizontal distance covered is

d=vt=(16 m/s)(2 s)=32 m

So, the closest answer is B.

5 0
3 years ago
How is the mass and speed of a particle related to its kinetic energy?
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Read 2 more answers
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