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slega [8]
3 years ago
9

One student did an experiment with two unknown minerals, Mineral 1 and Mineral 2. The hardness scale shown below was used for th

e experiment.
Mineral Talc Gypsum Calcite Fluorite Apatite Orthoclase
Hardness 1 2 3 4 5 6

Any mineral on the hardness scale would scratch minerals below it on the scale. The student tested Mineral 1 and observed that it could scratch only two other minerals on the hardness scale. It was observed that Mineral 2 could scratch only four other minerals on the hardness scale. Which of these is a correct conclusion the student can make about Mineral 1 and Mineral 2?
A) Mineral 1 can be used to scratch talc and Mineral 2 can be used to scratch orthoclase.
B) Mineral 1 can be used to scratch orthoclase and Mineral 2 can be used to scratch talc.
C) Mineral 1 can be used to scratch calcite and Mineral 2 can be used to scratch gypsum.
D) Mineral 1 can be used to scratch gypsum and Mineral 2 can be used to scratch calcite.
Physics
1 answer:
Umnica [9.8K]3 years ago
3 0
Scale: (soft 1-->6 hardest)
<span>1=Talc 2=Gypsum 3=Calcite 4=Fluorite 5=Apatite 6=Orthoclase

</span>Mineral #1 can only scratch two other minerals therefore it must have a hardness level of 2+1=3 which is Calcite. (scratches talc, gypsum)

Mineral #2 can scratch four other minerals therefore it must have a hardness level of 4+1=5 which is Apatite. (scratches all but Apatite, Orthoclase)

Looking through the possible conclusions.. It looks like answer is D.


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A playground merry-go-round has a mass of 115 kg and a radius of 2.50 m and it is rotating with an angular velocity of 0.520 rev
tatuchka [14]

Answer:

W_f = 2.319 rad/s

Explanation:

For answer this we will use the law of the conservation of the angular momentum.

L_i = L_f

so:

I_mW_m = I_sW_f

where I_m is the moment of inertia of the merry-go-round, W_m is the initial angular velocity of the merry-go-round, I_s is the moment of inertia of the merry-go-round and the child together and W_f is the final angular velocity.

First, we will find the moment of inertia of the merry-go-round using:

I = \frac{1}{2}M_mR^2

I = \frac{1}{2}(115 kg)(2.5m)^2

I = 359.375 kg*m^2

Where M_m is the mass and R is the radio of the merry-go-round

Second, we will change the initial angular velocity to rad/s as:

W = 0.520*2\pi rad/s

W = 3.2672 rad/s

Third, we will find the moment of inertia of both after the collision:

I_s = \frac{1}{2}M_mR^2+mR^2

I_s = \frac{1}{2}(115kg)(2.5m)^2+(23.5kg)(2.5m)^2

I_s = 506.25kg*m^2

Finally we replace all the data:

(359.375)(3.2672) = (506.25)W_f

Solving for W_f:

W_f = 2.319 rad/s

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