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Tpy6a [65]
3 years ago
6

A group of Advanced Open Water Divers plans to make two dives. The first dive is on a reef in 90 feet of water for 20 minutes. T

he group then remains on the surface for 1 hour. The second dive is on a wreck in 60 feet of water, with a planned bottom time of 30 minutes. What will be the ending pressure group after the second dive
Physics
1 answer:
Kaylis [27]3 years ago
5 0

Answer:

The ending pressure is R.

Explanation:

This is Recreational scuba diving (R). According to standards, this diving has prescribed depth limit of 130 ft. This means that the limit must include a depth that is not greater than 130 ft while using only compressed air and not even requiring a decompression stop.

Now, from the question, we can see that that there was a prescribed limit of 60ft of water with a planned bottom time of 30 minutes.

Thus, the ending pressure is R.

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Explain how the sun's position in the sky affects the length of shadows
Serggg [28]

A shadow forms on the side of an object that faces away from the sun. The length of shadows changes as Earth rotates. In the morning, the sun is low in the eastern sky and shadows are long. As time passes in the morning, the sun seems to move higher in the sky.

8 0
3 years ago
How do organisms use communication to survive?
just olya [345]

Answer: Im not entirly sure but I think It's D all the above. I think all but B because I never really heard of that but if you look in our history I think that happen im not sure I would wait untill you know that somone knows for sure.

Explanation:

3 0
2 years ago
On a straight, level, two-lane road, two cars moving in opposite directions approach and pass each other. Car A is in the eastbo
ludmilkaskok [199]

Answer:

a) 42 m/s, positive direction (to the east), b) 42 m/s, negative direction (to the west).

Explanation:

a) Let consider that Car A is moving at positive direction. Then, the relative velocity of Car A as seen by the driver of Car B is:

\vec v_{A/B} = \vec v_{A} - \vec v_{B}\\\vec v_{A/B} = 11 \frac{m}{s} \cdot i + 31 \frac{m}{s} \cdot i\\\vec v_{A/B} = 42 \frac{m}{s} \cdot i

42 m/s, positive direction (to the east).

b) The relative velocity of Car B as seen by the drive of Car A is:

\vec v_{B/A} = \vec v_{B} - \vec v_{A}\\\vec v_{B/A} = -31 \frac{m}{s} \cdot i - 11 \frac{m}{s} \cdot i\\\vec v_{B/A} = - 42 \frac{m}{s} \cdot i

42 m/s, negative direction (to the west).

5 0
3 years ago
A constant force of 8.0 N is exerted for 4.0 s on a 16-kg object initially at rest. The change in speed of this object will be:
AfilCa [17]

The change in speed of this object is 3m/s

According to Newton's second law;

F = ma

F = mv/t

Given the following parameters

Force F = 8.0N

mass m = 16kg

time t = 4.0s

Required

speed v

Substitute the given parameters into the formula

v = Ft/m

v = 8 * 6/16

v = 48/16

v = 3m/s

Hence the change in speed of this object is 3m/s

Learn more here: brainly.com/question/19072061

8 0
3 years ago
What is the moment of inertia of an object that rolls without slipping down a 3.5-m- high incline starting from rest, and has a
Daniel [21]

Answer:

I = 0.287 MR²

Explanation:

given,

height of the object = 3.5 m

initial velocity = 0 m/s

final velocity  = 7.3 m/s

moment of inertia = ?

Using total conservation of mechanical energy

change in potential energy will be equal to change in KE (rotational) and KE(transnational)

PE = KE(transnational) + KE (rotational)

mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}I\omega^2

v = r ω

mgh = \dfrac{1}{2}mv^2 + \dfrac{1}{2}\dfrac{Iv^2}{r^2}

I = \dfrac{m(2gh - v^2)r^2}{v^2}

I = \dfrac{mr^2(2\times 9.8 \times 3.5 - 7.3^2)}{7.3^2}

I =mr^2(0.287)

I = 0.287 MR²

3 0
3 years ago
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