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11 months ago

. A certain sample of metal has a mass of 3 kg and a volume of 250 cm3. What is its density?

Physics

1 answer:

LekaFEV [45]11 months ago

4 0

Given:

The mass of the sample is,

$m=3\text{ kg}$

The volume is

$V=250\text{ cm}^3$

To find:

The density

Explanation:

The volume is,

$\begin{gathered} V=250\text{ cm}^3 \\ =250\times10^{-6}\text{ m}^3 \end{gathered}$

The density is,

$\begin{gathered} d=\frac{m}{V} \\ =\frac{3}{250\times10^{-6}} \\ =12000\text{ kg/m}^3 \\ =12\times10^3\text{ kg/m}^3 \end{gathered}$

Hence, the density is

$12\times10^3\text{ kg/m}^3$

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The mechanical energy of this spacecraft is the sum of:

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On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

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Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

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Rearrange this equation to find an equation for $v$ :

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