Many of the boating fatalities take place after capsize, but a modest list of things to do before and after a capsize can minimize boat accidents and boat accident injuries.
Initially there is an significant list of thing to do before you even step on the boat:
1. Take the boat safety and water safely courses
2. Make certain that yourself and everyone else on the boat is wearing a well-fitting and safe life jacket.
3. Go over the place of the safety items with everyone on the boat as well as the location of the horn of the boat and the flare of the boat.
4. Paint bright color the hull of the boat in order to be seen easily from the air.
After a capsize, there are significant steps to make
1. Stay calm
2. Execute a head count and check everybody for injuries or immediate dangers.
3. Ensure that everyone has floatation device that coolers and other items that can be used.
4. Stay in the capsized boat unless dangerous.
5. Try to right the boat if someone has a knowledge on how to do so.
6. Use signal devices such as flares, bright colored life jacket, whistles, flashlights and mirror.
7. Try to reboard or climb onto it in order to get as much of your body out of the cold water as possible because treading water will ground to lose body heat sooner.
8. Do not waste energy and only signal when needed. Try to keep warm and stay strong<span />
Answer:
S = Vo t + 1/2 a t^2 distance traveled
t = (V2 - V1) / a = (0 - 21) / -3.5 = 6 sec time to stop
S = 21 * 6 - 3.5 * 6^2 / 2 = 63 m distance traveled
Responder:
Explicación:
Para que podamos calcular el tiempo que le tomará al esquiador permanecer en el aire, encontraremos el tiempo de vuelo como se muestra;
T = 2Usin theta / 2
theta = 90 grados
U = 25 m / s
T = 25sin90 / 2 (9,8)
T = 25 / 19,62
T = 1,27 segundos
Por lo tanto, los cielos usarán 1.27 segundos en el aire.
La distancia horizontal es el rango;
Rango R = U√2H / g
R = 25√2 (80) /9,8
R = 25√160 / 9,8
R = 25 * √16,326
R = 25 * 4.04
R = 101,02 m
Por tanto, la distancia horizontal recorrida por el esquiador es 101,02 m
Answer:
d²x/dt² = - 4dx/dt - 4x is the required differential equation.
Explanation:
Since the spring force F = kx where k is the spring constant and x its extension = 2.45 equals the weight of the 4 kg mass,
F = mg
kx = mg
k = mg/x
= 4 kg × 9.8 m/s²/2.45 m
= 39.2 kgm/s²/2.45 m
= 16 N/m
Now the drag force f = 16v where v is the velocity of the mass.
We now write an equation of motion for the forces on the mass. So,
F + f = ma (since both the drag force and spring force are in the same direction)where a = the acceleration of the mass
-kx - 16v = 4a
-16x - 16v = 4a
16x + 16v = -4a
4x + 4v = -a where v = dx/dt and a = d²x/dt²
4x + 4dx/dt = -d²x/dt²
d²x/dt² = - 4dx/dt - 4x which is the required differential equation
Answer:
<h2><u>Constant</u></h2>
Explanation:
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<h2>Thanks</h2>
