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netineya [11]
4 years ago
13

How are gems different from regular minerals? Gems are softer than regular minerals. Regular minerals are more beautiful than ge

ms. Gems are rarer than regular minerals. Regular minerals are less malleable than gems.
Physics
2 answers:
harkovskaia [24]4 years ago
8 0

Answer:

Option (3)

Explanation:

Gemstones are those pieces of minerals that are being cut and are in their polished form. They are economically valuable and the factors that control the value of these gemstones are color, cutting in perfect shapes, luster, durability, and rarity.

Whereas, regular minerals are those that occur naturally and are inorganic in nature. They are comprised of one or multiple elements. They are commonly found on earth.

From the given question, the best option to differentiate between regular minerals and gemstones is that gemstones are very rare and are economically very valuable, in comparison to the common regular minerals.

Thus, the correct answer is option (3).

geniusboy [140]4 years ago
6 0
And the answer is... C. A gem is typically much harder to find than a regular mineral because the are many more known sites of regular minerals being mined than there are of gems.
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3 years ago
9. A 227 kg object is moved a distance of 2.4 m forward by a force. If 686 J of work is done on the object, what is the object’s
Korolek [52]

<em>1</em><em>.</em><em>259ms^2</em>

Explanation:

since, WORK DONE = FORCE*DISTANCE

AND, FORCE=MASS*ACCELERATION

SO, THE WORK DONE BECOMES=MASS*ACCELERATION*DISTANCE

ACCELERATION=WORK/(MASS*DISTANCE)

AND, WORK=686J

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DISTANCE=2.4m

THEREFORE, ACCELERATION=686/(227*2.4)

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=1.259ms^2

4 0
3 years ago
Baseball scouts often use a radar gun to measure the speed of a pitch. One particular model of radar gun emits a microwave signa
Firdavs [7]

Since the Units presented are not in the International System we will proceed to convert them. We know that,

1 mi/h = 0.447 m/s

So the speed in SI would be

V=95mi/h(\frac{0.447m/s}{1mi/h})

V=42.465 m/s

The change in frequency when the wave is reflected is

f'=f(1+\frac{V}{c})

Or we can rearrange the equation as

f' = f + f\frac{V}{c}

f' = Apparent frequency

f = Original Frequency

c = Speed of light

f'-f = f\frac{V}{c}

\Delta f = f\frac{V}{c}

Replacing,

\Delta f = (10.525*10^9)(\frac{42.465}{3*10^8})

\Delta f =1489.8 Hz

Since the waves are reflected, hence the change in frequency at the gun is equal to twice the change in frequency

\Delta f_T = 2 \Delta f

\Delta f_T = 2(1489.8Hz)

\Delta f_T = 2979.63Hz

Therefore the increase in frequency is 2979.63Hz

4 0
4 years ago
A force of 44 N will stretch a rubber band 88 cm ​(0.080.08 ​m). Assuming that​ Hooke's law​ applies, how far will aa 11​-N forc
Setler79 [48]

Answer:

<em>The rubber band will be stretched 0.02 m.</em>

<em>The work done in stretching is 0.11 J.</em>

Explanation:

Force 1 = 44 N

extension of rubber band = 0.080 m

Force 2 = 11 N

extension = ?

According to Hooke's Law, force applied is proportional to the extension provided elastic limit is not extended.

F = ke

where k = constant of elasticity

e = extension of the material

F = force applied.

For the first case,

44 = 0.080K

K = 44/0.080 = 550 N/m

For the second situation involving the same rubber band

Force = 11 N

e = 550 N/m

11 = 550e

extension e = 11/550 = <em>0.02 m</em>

<em>The work done to stretch the rubber band this far is equal to the potential energy stored within the rubber due to the stretch</em>. This is in line with energy conservation.

potential energy stored = \frac{1}{2}ke^{2}

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3 0
3 years ago
Airbags and safety belts can reduce injuries because they can
pashok25 [27]

Answer:

reduce the velocity of collision

5 0
3 years ago
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