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iragen [17]
3 years ago
9

You are given five transparent objects: a calcite crystal, a diamond, a piece of window glass, a sample of quartz, and a piece o

f zircon. How would you go about identifying each sample?
Physics
1 answer:
blsea [12.9K]3 years ago
5 0

Answer:

Explanation:

The calcite Crystal can be identified by carrying out an acid test on it. This is done by bringing it in contact with a weak acid which cause crack in it structure. Hence a little of carbondioxide gas is released.

Since diamond is the hardest object,it can be identified on a Mohs scale. Likewise it can be tested by bringing it in contact with a newspaper, if the letters on the paper are not seen, that shows it is a pure diamond.

Window glass is identified by the code on it.

While Quartz Crystal is identified by scratch test. When it is used to scratch all other softer stones and metal, it leaves mark on them.

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The answer for question 2 i guess it’s c
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3 years ago
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Calculate the speed of an object that travels 75m in 15s
vodka [1.7K]

Answer:

speed =distance/time taken

5 m/s

6 0
3 years ago
The _______ are found on the right side of the arrow in a chemical reaction.
Dimas [21]
Reactants ---> product
The products are found on the right side of the arrow in a chemical reaction.
5 0
3 years ago
A fisherman is fishing from a bridge and is using a "50.0-N test line." In other words, the line will sustain a maximum force of
umka21 [38]

Answer:(a) 50 N

(b)38.34 N

Explanation:

Given

Maximum tension(T) in line 50 N

(a)If line is moving up with constant velocity i.e. there is no acceleration

This will happen when Tension is equal to weight of Fish

T-mg=0

T=mg

Maximum weight in this case will be 50 N

(b)acceleration of magnitude 2.98 m/s^2

T-mg=ma

50=m\left ( g+a\right )

50=m\left ( 12.78\right )

m=3.91

Therefore weight is 3.91\times 9.8=38.34 N

7 0
4 years ago
A 2.64-kg copper part, initially at 400 K, is plunged into a tank containing 4 kg of liquid water, initially at 300 K. The coppe
marin [14]

Answer:

a) T_f=305.7049\ K

b) \Delta S=313.51\ J.K^{-1}

Explanation:

Given:

  • mass of copper, m_c=2.64\ kg
  • initial temperature of copper, T_{ic}=400\ K
  • specific heat capacity of copper, c_c=385\ J.kg^{-1}.K^{-1}
  • mass of water, m_w=4\ kg
  • initial temperature of water, T_{iw}=300\ K
  • specific heat capacity of water, c_w=4200\ J.kg^{-1}.K^{-1}

a)

<u>∵No heat is lost in the environment and the heat is transferred only between the two bodies:</u>

Heat rejected by the copper = heat absorbed by the water

2.64\times 385\times (400-T_f)= 4\times 4200\times (T_f-300)

T_f=305.7049\ K

b)

<u>Now the amount of heat transfer:</u>

Q=m_c.c_c.(T_{ic}-T_{f})

Q=2.64\times 385\times (400-305.7049)

Q=95841.5841\ J

∴Entropy change

\Delta S=\frac{dQ}{T}

\Delta S=\frac{95841.5841}{305.7049}

\Delta S=313.51\ J.K^{-1}

5 0
4 years ago
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