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Pavlova-9 [17]
3 years ago
14

My buddy and I have just finished a dive to 15 metres/50 feet for 60 minutes. We want to return to the same site and depth and s

tay another 60 minutes. We can ______________________ to see about how long we have to remain at the surface to have enough no stop time. (choose all that apply)
Physics
1 answer:
marishachu [46]3 years ago
7 0

Answer:

1) Periodically check the no stop or NDL time on their computers

2) The dive computer planning mode can be used if available

3) Make use of a dive planning app

4) Check data from the RDP table or an eRDPML

Explanation:

The no stop times information from the computer gives the no-decompression limit (NDL) time allowable which is the time duration a diver theoretically is able to stay at a given depth without a need for a decompression stop

The dive computer plan mode or a downloadable dive planning app are presently the easiest methods of dive planning

The PADI RDP are dive planners based on several years of experience which provide reliable safety limits of depth and time.

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A plane travels 2.5 KM at an angle of 35 degrees to the ground, then changes direction and travels 5.2 km at an angle of 22 degr
Solnce55 [7]

Answer:

7.7 km 26°

Explanation:

The total x component is:

x = 2.5 cos(35°) + 5.2 cos(22°) = 6.87

The total y component is:

y = 2.5 sin(35°) + 5.2 sin(22°) = 3.38

The magnitude is:

d = √(x² + y²)

d = 7.7 km

The direction is:

θ = atan(y/x)

θ = 26°

5 0
3 years ago
What is the mass of something that weighs 300 N on earth
jeka57 [31]

Answer:

30.5810 kg

Explanation:

6 0
3 years ago
Mary starts from her house, walks 80 meters south, and stops to chat with her aunt on the sidewalk. After chatting for a few min
marin [14]
Mary walks:
d 1 = 80 m,  d 2 = 125 m,  d 3 = 45 m
t = 10 minutes = 600 seconds;
Average speed:
v = ( d 1 + d 2 + d 3 ) / t 
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v = 250 m / 600 s
v = 0.4167 m/s ≈ 0.42 m/s
Answer:
E ) 0.42 meters/second
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3 years ago
Which term describes the process of transferring charge without the direct contact
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Induction............
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A horizontal force of 14.0N is applied to a box of m=32.5kg with Vo=0. Ignoring friction, how far does the crate travel in 10.0s
Alex Ar [27]
I’m going to assume initial velocity is 0.

Use Newton’s second law:

F = m•a

F/m = a

14.0/32.5kg= 28/65 m/s^2

Use constant SUVAT acceleration formulae:

S- displacement - what we need to find out

U - initial velocity - 0

V

A - 28/65 m/s^2

T - 10 seconds

S = ut + 1/2at^2

Since u = 0

S = 1/2at^2

1/2• 28/65 • 10^2 = 21.5metres~

Answer is 21.5 metres

~Hoodini, here to help.
6 0
3 years ago
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