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3 years ago

What is Wave motion las ondas fisica movimiento ondulatorio

Physics

1 answer:

PIT_PIT [208]3 years ago

8 0

Answer:

Most familiar are surface waves on water, but both sound and light travel as wavelike disturbances, and the motion of all subatomic particles exhibits wavelike properties

Explanation:

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You tie the loose end of a 0.1 kg yo-yo string to your finger and then release the yo-yo so that it spins down toward the ground

Nikitich [7]

Answer:

The answer is " $5.06 \times 10^{-6} \ kg \ m^2$ "

Explanation:

$\to E_1=0..............(i)\\\\\to E_2= \frac{mV^2}{2} +\frac{Iw^2}{2} - mgh.............(ii)\\\\ \Delta E=0\\\\\to mgh= \frac{mV^2}{2} +\frac{Iw^2}{2} \\\\ \to 2 \ mgh= mV^2 +Iw^2\\\\ \to 2 \ mgh- mV^2 =Iw^2\\\\ \to m(2gh- V^2) =Iw^2\\\\ \to I= \frac{m(2gh- V^2)}{w^2}$

$= 5.06 \times 10^{-6} \ kg \ m^2$

3 0

3 years ago

A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 80.6 m/s at ground level.

kow [346]

Before the engines fail, the rocket's altitude at time t is given by

$y_1(t)=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

and its velocity is

$v_1(t)=80.6\dfrac{\rm m}{\rm s}+\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t$

The rocket then reaches an altitude of 1150 m at time t such that

$1150\,\mathrm m=\left(80.6\dfrac{\rm m}{\rm s}\right)t+\dfrac12\left(3.90\dfrac{\rm m}{\mathrm s^2}\right)t^2$

Solve for t to find this time to be

$t=11.2\,\mathrm s$

At this time, the rocket attains a velocity of

$v_1(11.2\,\mathrm s)=124\dfrac{\rm m}{\rm s}$

When it's in freefall, the rocket's altitude is given by

$y_2(t)=1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

where $g=9.80\frac{\rm m}{\mathrm s^2}$ is the acceleration due to gravity, and its velocity is

$v_2(t)=124\dfrac{\rm m}{\rm s}-gt$

(a) After the first 11.2 s of flight, the rocket is in the air for as long as it takes for $y_2(t)$ to reach 0:

$1150\,\mathrm m+\left(124\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=32.6\,\mathrm s$

So the rocket is in motion for a total of 11.2 s + 32.6 s = 43.4 s.

(b) Recall that

${v_f}^2-{v_i}^2=2a\Delta y$

where $v_f$ and $v_i$ denote final and initial velocities, respecitively, $a$ denotes acceleration, and $\Delta y$ the difference in altitudes over some time interval. At its maximum height, the rocket has zero velocity. After the engines fail, the rocket will keep moving upward for a little while before it starts to fall to the ground, which means $y_2$ will contain the information we need to find the maximum height.

$-\left(124\dfrac{\rm m}{\rm s}\right)^2=-2g(y_{\rm max}-1150\,\mathrm m)$

Solve for $y_{\rm max}$ and we find that the rocket reaches a maximum altitude of about 1930 m.

(c) In part (a), we found the time it takes for the rocket to hit the ground (relative to $y_2(t)$ ) to be about 32.6 s. Plug this into $v_2(t)$ to find the velocity before it crashes: