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3 years ago

15

"Today you will use this chart to help you identify some common minerals," instructed Mr. Grant, Jimmy's teacher. Each group of

students was given three mineral samples to identify. Jimmy and his partners were puzzled by one sample. It scratched glass, so it was pretty hard. One student thought it was feldspar; another thought it was quartz. What other properties could the group use to identify the mineral?

2 answers:

Zigmanuir [339]3 years ago

7 0

The correct answer to the question is luster and color

Brainliest plz!

scoray [572]3 years ago

4 0

The properties of minerals that might help Jimmy and his partners to identify the sample other than Hardness (scratch test) are:

1. Color

2. Crystalline Structure

3. Cleavage or Fracture

4. Transparency

5. Tenacity

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Answer:

$t'=1.1897*10^{-6} s$

t'=1.1897 μs

Explanation:

First we will calculate the velocity of micrometeorite relative to spaceship.

Formula:

$u=\frac{u'+v}{1+\frac{u'*v}{c^{2}}}$

where:

v is the velocity of spaceship relative to certain frame of reference = -0.82c (Negative sign is due to antiparallel track).

u is the velocity of micrometeorite relative to same frame of reference as spaceship = .82c (Negative sign is due to antiparallel track)

u' is the relative velocity of micrometeorite with respect to spaceship.

In order to find u' , we can rewrite the above expression as:

$u'=\frac{v-u}{\frac{u*v}{c^{2} }-1 }$

$u'=\frac{-0.82c-0.82c}{\frac{0.82c*(-0.82c)}{c^{2} }-1 }$

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Answer:

The right answer is:

(a) 63.83 kg

(b) 0.725 m/s

Explanation:

The given query seems to be incomplete. Below is the attachment of the full question is attached.

The given values are:

T = 3 sec

k = 280 N/m

(a)

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or,

⇒ $m=\frac{k T^2}{4 \pi^2}$

On substituting the values, we get

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(b)

The speed of the string will be:

⇒ $\frac{1}{2}k(0.4)^2=\frac{1}{2}k(0.2)^2+\frac{1}{2}mv^2$

then,

⇒ $v=\sqrt{\frac{k((0.4)^2-(0.2)^2)}{m} }$

On substituting the values, we get

⇒ $=\sqrt{\frac{280\times ((0.4)^2-(0.2)^2)}{63.83} }$

⇒ $=\sqrt{\frac{280(0.16-0.04)}{63.83} }$

⇒ $=\sqrt{\frac{280\times 0.12}{63.83} }$

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