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3 years ago

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An engineer has the task of producing an aluminum alloy with a density of 3.0 grams per cubic centimeter. She comes up with the

perfect alloy and saves a sample of it. Which mass and volume measurements correspond to this sample?
7 g and 2.3 cm³
10 g and 7 cm³
15 g and 5 cm³
21 g and 8 cm³

1 answer:

pochemuha3 years ago

5 0

Answer:

The best option is for the following option m = 15 [g] and V = 5 [cm³]

Explanation:

We have that the density of a body is defined as the ratio of mass to volume.

$Ro =m/V$

where:

Ro = density = 3 [g/cm³]

Now we must determine the densities with each of the given values.

<u>For m = 7 [g] and V = 2.3 [cm³]</u>

$Ro=7/2.3\\Ro=3.04 [g/cm^{3} ]$

<u>For m = 10 [g] and V = 7 [cm³]</u>

<u /> $Ro=10/7\\Ro=1.42[g/cm^{3} ]\\$ <u />

<u>For m = 15 [g] and V = 5 [cm³]</u>

<u /> $Ro=15/5\\Ro=3[g/cm^{3} ]\\$ <u />

<u>For m = 21 [g] and V = 8 [cm³]</u>

<u /> $Ro=21/8\\Ro=2.625[g/cm^{3} ]\\$ <u />

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Given:

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Distance traveled (s) = ?

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v = u + at

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Answer:

$v_{max}=52.38\frac{m}{s}$

$v_{100}=33.81$

Explanation:

the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:

$\sum{F}=0=F_d-W$

$F_d=W$

$kv_{max}^2=m*g$

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To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:

$\sum{F}=ma=W-F_d$

$ma=W-F_d$

$ma=mg-kv_{100}^2$

$a=g-\frac{kv_{100}^2}{m}$ (1)

consider the next equation of motion:

$a = \frac{(v_{x}-v_0)^2}{2x}$

If assuming initial velocity=0:

$a = \frac{v_{100}^2}{2x}$ (2)

joining (1) and (2):

$\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}$

$\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g$

$v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g$

$v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}$

$v_{100}=\sqrt{\frac{g}{(\frac{1}{2x}+\frac{k}{m})}}$ (3)

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}$

$v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}$

$v_{100}=\sqrt{1,143.3}$

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To plot velocity as a function of distance, just plot equation (3).

To plot velocity as a function of time, you have to consider the next equation of motion:

$v = v_0 +at$

as stated before, the initial velocity is 0:

$v =at$ (4)

joining (1) and (4) and reducing you will get:

$\frac{kt}{m}v^2+v-gt=0$

solving for v:

$v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }$

Plots:

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3 years ago