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3 years ago

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Solve the problem using dimensional analysis. Every number must have a unit. Work must be shown. Conversion factors are in the t

able.
Gold cost $8.00 per ounce. Determine how many grams you can purchase for $15.00.

1 answer:

anygoal [31]3 years ago

5 0

The number of grams of gold you can purchase for $15 if Gold cost $8.00 per ounce is 53.26g

Dimensional analysis is a means of representing values using units.

From the question, we are told that Gold cost $8.00 per ounce, this means that:

$8.00 = 1 ounce

In order to calculate the amount in grams you can purchase for $15, we need to know the price in ounces first.

$15.00 = x

Divide both expressions

$\dfrac{8}{15} = \dfrac{1}{x}\\x = \dfrac{15}{8}\\x= 1.875 ounces$

Next is to convert the ounces to pounds.

1 pound = 16 ounces

y = 1.875 ounces

y = 1.875/16

y = 0.1171875 pounds

Conver 0.1171875 pounds to kg

Since 1kg = 2.2 pounds

z = 0.1171875 pounds

2.2z = 0.1171875

z = 0.1171875/2.2

z = 0.053267kg

Finally, convert 0.053267kg to grams using the conversion rate

1kg = 1000g

0.053267kg = y

y = 0.05326 * 1000

y = 53.26g

This shows that the amount of grams you can purchase for $15 is 53.26g

Learn more here: brainly.com/question/11819069

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4 years ago

An object is located in air, 25 cm from the vertex of the concave surface of a block of glass (as viewed from the air side of th

Marizza181 [45]

Your question is missing one part as "magnification"

i have completed the missing part below

Answer:

a. $d_{i}=-0.0566cm$

b. $M=441.69$

Explanation:

For this type of numerical we will use the following formulas

$\frac{n1}{d_{o} }+\frac{n_{2} }{d_{i} }=\frac{n_{2}-n_{1} }{R}$ ......... Eq1

where,

$n_{1}$ $=$ refractive index of the medium surrounding refracting surface/object

i.e. $n_{1}$ = $n_{air}$

$n_{2}$ $=$ refractive index of the refracting surface/object

i.e. $n_{2}=n_{glass}$

$d_{0}=$ distance of object from the vertex of the refracting surface

$d_{i}=$ distance of image from the vertex of the refracting surface

$R=$ radius of curvature of the refracting surface

$M=\frac{d_{0} }{d_{i} }$ ........... Eq2

where,

$M=$ magnification

Convention:

$R>0\for\ the\ convex\ refractive\ surface\ of\ curvature\\\ R$

Given:

$n_{1}$ = $n_{air}$ $=1.0$

$n_{2}=n_{glass}$ $=1.5$

$d_{0}=$ $25cm$

$R=-11$ $cm$ because refraction surface is concave

Required:

a. $d_{i}=$ $?$

b. $M=?$

Solution:

a. putting values in eq1, we get

$\frac{1.0}{25}+\frac{1.5}{d_{i} }=\frac{1.5-1.0}{-11}$

$\frac{1.5}{d_{i} }=-0.045-0.040$

$d_{i}=(-0.085)(\frac{1}{1.5} )$

$d_{i}=-0.0566$ cm

b. $M=\frac{25}{-0.05665}$

$M=441.69$

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4 years ago

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Hitman42 [59]

Answer:

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Explanation:

given,

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Since the vessel is moving at a constant speed, the resultant force on it should equal zero.

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Mass of water,

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