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3 years ago

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A golden-colored cube is handed to you. The person wants you to buy it for $100, saying that is a gold nugget. You pull out your

old geology text and look up gold in the mineral table, and read that its density is 19.3 g/cm3. You measure the cube and find that it is 2 cm on each side, and weighs 40 g. What is its density? Is it gold? Should you buy it?

2 answers:

kenny6666 [7]3 years ago

6 0

Answer:

Explanation:

Volume of cube=side³

=2*2*2

=8 cm³

Mass of cube=40g

Its density=mass/volume

=5g/cm³

which differs the value given in the text hence it is not gold and u should not bye it...

Instead u should call the police for arresting the person lol...

romanna [79]3 years ago

3 0

Don’t buy it it’s not gold :)

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The ice skaters partner liftes her up a distance of 1 m work done or not work done

SOVA2 [1]

Answer:

Work done.

Explanation:

The skater who lifts has to overcome the partner's weight. When lifted up by 1 meter, her potential energy increases by (mass)x(gravitational acceleration)x(1meter), which is the amount of work done.

(This all assumes lifting vertically and no other forces being part of the picture)

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3 years ago

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

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3 years ago

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STatiana [176]

A. attract each other.

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3 years ago

g A 1.45-kg block is pushed against a vertical wall by means of a spring (k = 860 N/m). The coefficient of static friction betwe

olga_2 [115]

Answer:

The minimum compression is $x= 0.046m$

Explanation:

From the question we are told that

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For the the block not slip it mean the sum of forces acting on the horizontal axis is equal to the forces acting on the vertical axis

Now the force acting on the vertical axis is the force due to gravity which is mathematically given as

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And the force acting on the horizontal axis is force due to the spring which is mathematically represented as

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Now equating this two formulas and making x the subject

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substituting values we have

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Savatey [412]

Answer:

B and C!

Explanation:

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