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swat32
3 years ago
9

How do noise and vibration affect you when operating a boat?

Physics
2 answers:
natta225 [31]3 years ago
4 0

Answer:

It will slow down your reaction time, and reducing ability to control the boat and it will give physical fatigue .

Explanation:

When you aboard a boat you are subjected to both low and high frequencies from the wind and from the engine of the boat. You can feel the waves which are of low frequency as a vibration.

Like other machines boat can produce noise and vibration together that will make you tired, and in some cases your reaction times will slow down and you will loose the ability to control that boat.

After that the heat and glare from the sun will also make you physically fatigue and in this condition operating a boat is more dangerous.

kogti [31]3 years ago
3 0
The noise and vibration could affect your body and mood in many ways while operating a boat. The noise and vibration could make you moody or uncomfortable, because it will also distract you in doing what you're supposed to do. Your body will also be affected, making your body tired because of the things surrounding you and the noise and vibration, it could affect your body in many ways.
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An object is originally moving at a constant velocity of 8 m/s in the -x direction. It moves at this constant velocity for 3 sec
aivan3 [116]

Answer:

244.64m

Explanation:

First, we find the distance traveled with constant velocity. It's simply multiplying velocity time the time that elapsed:

x = V*t = -8\frac{m}{s} *3s = -24m

After this, the ball will start traveling with a constant acceleration motion. Due to the fact that the acceleration is the opposite direction to the initial velocity, this motion will have 2 phases:

1. The velocity will start to decrease untill it reaches 0m/s.

2. Then, the velocity will start to increase at the rate of the acceleration.

The distance that the ball travels in the first phase can be found with the following expression:

v^2 = v_0^2 + 2a*d

Where v is the final velocity (0m/s), v_0 is the initial velocity (-8m/s) and a is the acceleration (+9m/s^2). We solve for d:

d = \frac{v^2 - v_0^2}{2a} = \frac{(0m/s)^2 - (-8m/s)^2}{2*7m/s^2}= -4.57m

Now, before finding the distance traveled in the second phase, we need to find the time that took for the velocity to reach 0:

t_1 = \frac{v}{a} = \frac{8m/s}{7m/s^2} = 1.143 s

Then, the time of the second phase will be:

t_2 = 9s - t_1 = 9s - 1.143s = 7.857s

Using this, we using the equations for constant acceleration motion in order to calculate the distance traveled in the second phase:

x = \frac{1}{2}a*t^2 + v_0*t + x_0

V_0, the initial velocity of the second phase, will be 0 as previously mentioned. X_0, the initial position, will be 0, for simplicity:

x = \frac{1}{2}*7\frac{m}{s^2}*t^2 + 0m/s*t + 0m = 216.07m

So, the total distance covered by this object in meters will be the sum of all the distances we found:

x_total = 24m + 4.57m + 216.07m = 244.64m

8 0
2 years ago
I have no idea of how to approach this problem
nataly862011 [7]

Answer:

p=1

Explanation:

Well me know that v=m/s

and that a=m/s^2

so

(m/s)^{2} = (m/s^2)(x^p)\\\\(m^2/s^2)/(m/s^2)=x^p\\\\m^2s^2/ms^2=x^p\\\\(m^2/m)(s^2/s^2)=x^p\\\\m=x^p\\\\p=1

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3 years ago
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3 years ago
A projectile is fired with an initial speed of 40 m/s at an angle of elevation of 30∘. Find the following: (Assume air resistanc
BabaBlast [244]

Answer:

a. 2.0secs

b. 20.4m

c. 4.0secs

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e. 40m/s, ∅= -30°

Explanation:

The following Data are giving

Initial speed U=40m/s

angle of elevation,∅=30°

a. the expression for the time to attain the maximum height is expressed as

t=\frac{usin\alpha }{g}

where g is the acceleration due to gravity, and the value is 9.81m/s if we substitute values we arrive at

t=40sin30/9.81\\t=2.0secs

b. the expression for the maximum height is expressed as

H=\frac{u^{2}sin^{2}\alpha  }{2g} \\H=\frac{40^{2}0.25 }{2*9.81} \\H=20.4m

c. The time to hit the ground is the total time of flight which is twice the time to reach the maximum height ,

Hence T=2t

T=2*2.0

T=4.0secs

d. The range of the projectile is expressed as

R=\frac{U^{2}sin2\alpha}{g}\\R=\frac{40^{2}sin60}{9.81}\\R=141.2m

e. The landing speed is the same as the initial projected speed but in opposite direction

Hence the landing speed is 40m/s at angle of -30°

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3 years ago
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