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4 years ago

14

An object accelerates to a velocity of 230 m/s over a time of 2.5 s. The acceleration it experienced was 42 m/s2. What was its i

nitial velocity?

1 answer:

zmey [24]4 years ago

5 0

Answer:

230 = x + 105

x= 125

Explanation:

v = v0 + at

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The rocket that sent us to the moon traveled at 25,000 mi hr convert to miles per second

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It’s the same it’s 25000

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If the density of pure copper is 8.89g/cm3(at 20•c),calculate the volume of a copper brick with a mass of 4.5kg

oksano4ka [1.4K]

The volume of a substance can directly be calculated by the ratio of mass and density. That is:

volume = mass / density

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Eleven wa=eighs 47 kg. her height is 1.63 m. what is her bmi

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Answer:

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3 years ago

Completenlo por favor

Ainat [17]

La velocidad $\mathbf v$ del objeto al tiempo $t$ es descrito por

$\mathbf v(t)=v_x(t)\,\mathbf i+v_y(t)\,\mathbf j$

donde

$\begin{cases}v_x(t)={v_x}_0+a_xt\\v_y(t)={v_y}_0+a_yt\end{cases}$

El objeto no tiene aceleración horizontal, pues $a_x=0$ y $v_x$ está determinado exactamente por su velocidad inicial en la dirección horizontal. En la dirección vertical, la gravedad es la única fuerza que actúa en el objeto, pues $a_y=-9.81\,\dfrac{\mathrm m}{\mathrm s^2}$ . Entonces, la velocidad después de $3\,\mathrm s$ satisface

${v_x}_0\,\mathbf i+\left({v_y}_0+\left(-9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)(3\,\mathrm s)\right)\,\mathbf j=20\,\mathbf i-4\,\mathbf j$

Inmediatamente, vemos que ${v_x}_0=20\,\dfrac{\mathrm m}{\mathrm s}$ y podemos encontrar que ${v_y}_0=25.43\,\dfrac{\mathrm m}{\mathrm s}$ .

Su posicíon es descrita al tiempo $t$ por

$\mathbr r(t)=r_x\,\mathbf i+r_y\,\mathbf j$

con

$\begin{cases}r_x={r_0}_x+{v_0}_xt+\dfrac12a_xt^2\\\\r_y={r_0}_y+{v_0}_yt+\dfrac12a_yt^2\end{cases}$

Si la posición inicial del objeto es el origen, suponemos que $\mathbr r(0)=\mathbf 0$ , y además tenemos

$\mathbf r(t)=\left(20\,\dfrac{\mathrm m}{\mathrm s}\right)t\,\mathbf i+\left(\left(25.43\,\dfrac{\mathrm m}{\mathrm s}\right)t+\dfrac12\left(-9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2\right)\,\mathbf j$

Queremos determinar el máximo de $r_y$ . Encontramos que ${r_y}_{\mathrm{max}}\approx32.9\,\mathrm m$ cuando $t\approx2.59\,\mathrm s$ .

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When your head is submerged your skull also vibrates with the sound because it is close to the same density and elasticity as water. Below the surface sound wave pass directly through water and into your head. And above the surface the sound waves only vibrates your ear drum unless the sound is very loud.

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