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anyanavicka [17]

3 years ago

11

What is most likely to cause someone to fall overboard? balancing the weight of people and gear sitting on a pedestal chair whil

e anchored fishing from a boat standing up in a boat?

1 answer:

Salsk061 [2.6K]3 years ago

6 0

The answer for this question is standing up in a boat.

Standing up in a boat is most likely to cause someone to fall overboard especially if someone starts to rock the boat or if a wave occurs and the person standing is not prepared of the said situations. You may also consider the causes standing on a slippery surface and slip off as well.

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A target behavior is

crimeas [40]

Answer:

Target Behavior would be the behavior known to be changed, it can be defined by function or topography

3 0

2 years ago

A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an

Nat2105 [25]

Answer:

a) $T=0.0031416s$

b) $v_{max}=6.283\frac{m}{s}$

c) $E=0.1974J$

d) $F=80N$

e) $F=40N$

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by $X=Acos(\omega t +\phi)$ (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

$\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)$ (2)

For the acceleration

$\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)$ (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

$A\omega^2=8x10^{3}\frac{m}{s^2}$

Since we know the amplitude A=0.002m we can solve for $\omega$ like this:

$\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}$

And we with this value we can find the period with the following formula

$T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s$

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

$v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}$

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

$E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J$

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens $cos(\omega t +\phi)=1$ , and we also know that the maximum acceleration is given by::

$|\frac{d^2X}{dt^2}|=A\omega^2$

So then we have that:

$F=ma=mA\omega^2$

And since we have everything we can find the force

$F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N$

6) Part e

When the mass it's at the half of it's maximum displacement the term $cos(\omega t +\phi)=1/2$ and on this case the acceleration would be given by;

$|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}$

And the force would be given by:

$F=ma=\frac{1}{2}mA\omega^2$

And replacing we have:

$F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N$

8 0

2 years ago

The inner solar system contains the

Dominik [7]

Jovian planets and dwarf planets like pluto are in the outer part of our solar system. so, terrestrial planets ,which means earth like planets, live in the inner part of our solar system. or B. is the answer

3 0

3 years ago

_____ has a longer wavelength than _____.

victus00 [196]

Answer:

B

Explanation:

its b ok ok bye

7 0

2 years ago

ANSWER One end of a string is attached to an object of mass M, and the other end of the string is secured so that the object is

qaws [65]

Answer:

Explanation:

It is a case of oscillation by simple pendulum . Expression for simple pendulum is given as follows

T = $2\pi\sqrt{\frac{l}{g} }$

where T is time period , l is length of pendulum and g is acceleration due to gravity .

$\frac{1}{f} =2\pi\sqrt{\frac{l}{g} }$ , f is frequency of oscillation

For the given case

$\frac{1}{f_o} =2\pi\sqrt{\frac{l}{g} }$

subsequently length becomes half so

$\frac{1}{f} =2\pi\sqrt{\frac{l}{2g} }$

dividing

$\frac{f}{f_o} = \sqrt{\frac{2}{1} }$

f = $\sqrt{2} f_o$

frequency of oscillation becomes √2 times.

4 0

3 years ago