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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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At what distance from Earth does the force of gravity exerted by Earth on the coasting spacecraft cancel the force of gravity ex
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346 * 10⁶ m

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The force of gravity of the earth that will cancel the the force of gravity exerted by the moon will be equal to each other

Let F_{e} be the force of gravity exerted by the earth

and let F_{m} be the force of gravity exerted by the moon

According to Newton's law of universal gravitation, the force of attraction between two different masses, m₁ and m₂ separated by a distance, d,  is given by:

F = \frac{Gm_{1} m_{2} }{d^{2} }

Mass of the earth, m_{e} = 5.97 * 10^{24} kg

Mass of the moon, m_{m} = 7.348 * 10^{22} kg

Mass of the satellite, m_{s} = ?

F_{e}  = \frac{G*5.97 * 10^{24} M }{d^{2} }...............................(1)

The earth and the moon are separated by a distance, 3.844 * 10⁸ m

F_{m}  = \frac{G*7.348 * 10^{22} M }{(3.844 * 10^{8} - d) ^{2} }............................(2)

Equating equations (1) and (2)

\frac{5.97 * 10^{24} }{d^{2} } = \frac{7.348 * 10^{24} }{(3.844* 10^{8} -d)^{2} }

(5.97 * 10^{24})(14.78 * 10^{16}  -7.688*10^{8}d + d^{2}) = 7.348 * 10^{24} d^{2} \\88.24*10^{40} - 45.9 * 10^{32}d +  5.97 * 10^{24}d^{2} =  7.348 * 10^{24} d^{2}\\ 1.378 * 10^{24}d^{2} + 45.9 * 10^{32}d + 88.24*10^{40} = 0\\

Factorising out 10^{24}

1.378d^{2} + 45.9 * 10^{8}d + 88.24*10^{16} = 0

Solving for d in the quadratic equation  above:

d = 346 * 10⁶ m

4 0
3 years ago
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