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3 years ago

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An industrial engineer is designing a process to manufacture bullets. The mass of each bullet must be within 0.25% of 170.0 grai

ns. What range of bullet masses, in mg, will meet this tolerance? (1 gr = 64.79891 mg)

1 answer:

kondaur [170]3 years ago

7 0

Answer:

The range is 11015.81468 ± 27.53953675 mg.

Explanation:

The mass of each bullet must be within 0.25% of 170.0 grains.

We need to calculate the range of bullet masses

Mass of grains is given by,

$\dfrac{0.25}{100}\times170.0=0.425$

$1\ g grain = 64.79891\ mg$

$0.425\ grains = 64.79891\times0.425$

$0.425\ grains = 27.53953675\ mg$

$170.0\ grains =\dfrac{27.5395367\times170.0}{0.425}$

$170.0\ grains =11015.81468\ mg$

Hence, The range of bullet masses are 11015.81468 ± 27.53953675 mg.

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Answer:

$\rho_{s} = 4$

$\rho_{l} = 0.6$

$\rho{liq} = 600 kg/m^{3}$

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Weight of solid in air, $w_{sa} = 200 N$

Weight of solid in water, $w_{sw} = 150 N$

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Solution:

Calculation of:

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$\rho_{s} = \frac{w_{sa}}{w_{sa} - w_{sw}}$

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2. Relative density of liquid, $\rho_{l}$

$\rho_{l} = \frac{w_{sa} - w_{sl}}{w_{sa} - w_{sw}}$

$\rho_{l} = \frac{200 - 170}{200 - 150} = 0.6$

3. Density of liquid in S.I units:

Also, we know:

$\rho{l} = \frac{\rho_{liq}}{\rho_{w}}$

where

$= {\rho_{liq}}$ = density of liquid

$= {\rho_{w}} = 1000 kg/m^{3}$ = density of water

Now, from the above formula:

$0.6 = \frac{\rho_{liq}}{1000}$

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