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?
1 answer:
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:
- 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).
- You'll check and keep the letter in that 47 row (S) for future use.
- 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).
- 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:
a) E = -4 10² N / C , b) x = 0.093 m, c) a = 10.31 m / s², θ=-71.9⁰
Explanation:
For that exercise we use Newton's second Law, in the attached we can see a free body diagram of the ball
X axis
-
= m a
Axis y
- W = 0
Initially the system is in equilibrium, so zero acceleration
Fe =
T_{y} = W
Let us search with trigonometry the components of the tendency
cos θ = T_{y} / T
sin θ = / T
T_{y} = cos θ
= T sin θ
We replace
q E = T sin θ
mg = T cosθ
a) the electric force is
= q E
E = / q
E = -0.032 / 80 10⁻⁶
E = -4 10² N / C
b) the distance to this point can be found by dividing the two equations
q E / mg = tan θ
θ = tan⁻¹ qE / mg
Let's calculate
θ = tan⁻¹ (80 10⁻⁶ 4 10² / 0.01 9.8)
θ = tan⁻¹ 0.3265
θ = 18 ⁰
sin 18 = x/0.30
x =0.30 sin 18
x = 0.093 m
c) The rope is cut, two forces remain acting on the ball, on the x-axis the electric force and on the axis and the force gravitations
X axis
= m aₓ
aₓ = q E / m
aₓ = 80 10⁻⁶ 4 10² / 0.01
aₓ = 3.2 m / s²
Axis y
W = m
a_{y} = g
a_{y} = 9.8 m/s²
The total acceleration is can be found using Pythagoras' theorem
a = √ aₓ² + a_{y}²
a = √ 3.2² + 9.8²
a = 10.31 m / s²
The Angle meet him with trigonometry
tan θ = a_{y} / aₓ
θ = tan⁻¹ a_{y} / aₓ
θ = tan⁻¹ (-9.8) / 3.2
θ = -71.9⁰
Movement is two-dimensional type with acceleration in both axes