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tatiyna
3 years ago
6

I am planning to make two dives. The first dive is to 60 feet for 45 minutes, and the second dive is to 60 feet for 35 minutes.

How long would I have to stay on the surface (minimum surface interval) to do these two dives safely?

Physics
1 answer:
Tpy6a [65]3 years ago
4 0

Answer:

1:04-1:10 hours

Explanation:

You'll need a <em>Recreational dive planner</em> table, I annexed a copy, now you'll follow the next steps:

  1. In the first part of your table, you'll look for the distance row (in feet) of your first dive, for this specific exercise you'll find 60, once you locate it you'll go down that column until you reach the time you'll dive, in this case, 45 (minutes) or the closest value (47).
  2. You'll check and keep the letter in that 47 row (S) for future use.
  3. Now you have to go to the second part of your table and look for the distance column, in feet, of your second dive. We find 60 and then going right in the blue row, we'll look for the time (35) or its closest value (36).
  4. Finally, we have to check the letter for 36 minutes (F) and we'll make it met with the letter S in the first portion of your tables. This will give us an interval of time, 1:04-1:10 in this case.

I hope you find this information useful and interesting! Good luck!

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Answer:

The momentum of the bowling ball is 53.4 kg-m/s.

Explanation:

We have,

Mass of a bowling ball is 8.9 kg

Speed of the ball is 6 m/s

It is required to find the momentum of the ball. The momentum of an object is given in terms of its mass and speed as :

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A series RLC circuit consists of a 52.0 Ω resistor, a 4.80 mH inductor, and a 330 nF capacitor. It is connected to an oscillator
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Answer:

(G) 75.11 ohm

(H) 0.08 A

(I) 46.2 degree

Explanation:

R = 52 ohm

L = 4.8 m H = 4.8 x 106-3 H

C = 330 nF = 330 x 10^-9 F

Vo = 6 V

(G)

f = 5000 Hz

Let the impedance is Z.

X_{L}= 2 \pi fL = 2 \times 3.14\times 5000\times 4.8\times 10^{-3}=150.72 ohm

X_{c}= \frac{1}{2 \pi fC}=\frac{1}{2\times 3.14\times 5000\times 330\times 10^{-9}}=96.51 ohm

Z=\sqrt{R^{2}+\left ( X_{L}-X_{c} \right )^{2}}

Z=\sqrt{52^{2}+\left (150.72-96.51)^{2}}=75.11 ohm

(H) Let Io be the peak current

I_{0}=\frac{V_{0}}{Z}=\frac{6}{75.11}=0.0798 A = 0.08 A

(I) Let Ф be the phase angle

tan\phi = \frac{X_{L}-X_{C}}{R}

tan\phi =\frac{150.72-96.51}}{52}=1.0425

Ф = 46.2 degree

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