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AlexFokin [52]
3 years ago
10

The density of gold is 19.3. G/cm(3). If a nugget of iron pyrite and nugget of gold each have a mass of 50 g, what can you concl

ude about the volume of each nugget
Physics
1 answer:
tangare [24]3 years ago
5 0
The pyrite will be bigger, because its density is much lower.

I <em>do</em> know that the gold's volume will be 2.5906 (With a bunch more numbers after it)

50 divided by 19.3 = 2.5906
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Risk of return on investment is higher than other forms of energy generation.
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A value with magnitude only is a ?​
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A superhero swings a magic hammer over her head in a horizontal plane. The end of the hammer moves around a circular path of rad
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Answer:

9.21954 m/s

54 m/s²

Angle is zero

Explanation:

r = Radius of arm = 1.5 m

\omega = Angular velocity = 6 rad/s

The horizontal component of speed is given by

v_h=\omega r\\\Rightarrow v_h=6\times 1.5\\\Rightarrow v_h=9\ m/s

The vertical component of speed is given by

v_v=2\ m/s

The resultant of the two components will give us the velocity of hammer with respect to the ground

v=\sqrt{v_h^2+v_v^2}\\\Rightarrow v=\sqrt{9^2+2^2}\\\Rightarrow v=9.21954\ m/s

The velocity of hammer relative to the ground is 9.21954 m/s

Acceleration in the vertical component is zero

Net acceleration is given by

a_n=a_h=\omega^2r\\\Rightarrow a_n=6^2\times 1.5\\\Rightarrow a_n=54\ m/s^2

Net acceleration is 54 m/s²

As the acceleration is towards the center the angle is zero.

3 0
3 years ago
A paperweight consists of a 8.55-cm-thick plastic cube. Within the plastic a thin sheet of paper is embedded, parallel to opposi
ella [17]

Answer:

\mu=1.5322

Explanation:

Given:

Thickness of the paperweight cube, x=8.55\ cm

apparent depth from one side of the inbuilt paper in the plastic cube, i=4.02\ cm

apparent depth from the other side of the inbuilt paper in the plastic cube, i'=1.56\ cm

Now as we know that refractive index is given as:

\rm \mu=\frac{real\ depth}{apparent\ depth}

  • Let the real depth form first side of the slab be, d
  • Then the depth from the second side of the slab will be, d'=x-d=8.55-d

Since refractive index for an amorphous solid is an isotropic quantity so it remains same in all the direction for this plastic.

\mu=\mu'

\frac{d}{i} =\frac{d'}{i'}

\frac{d}{4.02}=\frac{8.55-d}{1.56}

d=6.1597\ cm

Now the refractive index:

\mu=\frac{d}{i}

\mu=\frac{6.1596}{4.02}

\mu=1.5322

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