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4 years ago

8

And metamorphic rock is classified according to its

2 answers:

ICE Princess25 [194]4 years ago

8 0

The answer is basis of texture

Lubov Fominskaja [6]4 years ago

7 0

Answer: A <span>metamorphic rock is classified according to its </span><span>basis of texture and mineral composition.

Hope this helps =)
Hava a nice day!

</span>

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Two technicians are discussing the testing of a catalytic converter. Technician A says that a vacuum gauge can be used and obser

Andrej [43]

Answer:

Answer is C. Both technicians A and B.

Refer below.

Explanation:

Two technicians are discussing the testing of a catalytic converter. Technician A says that a vacuum gauge can be used and observed to see if the vacuum drops with the engine at 2500 RPM for 30 seconds. Technician B says that a pressure gauge can be used to check for backpressure. The following technician is correct:

Both technicians A and B

7 0

3 years ago

Read 2 more answers

What is the mass of 5.4 moles of copper ?

Helen [10]

343.1484 grams. this is your answer.

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4 years ago

How many stars are in the universe (approximately)? O 40 sixtillion 0 365 billion O 86.4 million O one

Annette [7]

Answer:

a i belive

Explanation:

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3 years ago

Gold is the most ductile of all metals. For example, one gram of gold can be drawn into a wire 2.05 km long. The density of gold

arsen [322]

Answer:

Resistance of gold wire, $R=1977 \times 10^3 ohm$

Explanation:

In this question we have given

Density of gold, $d=19.3\times 10^3 \frac{kg}{m^3}$

resistivity of gold, $r=2.44\times 10^{-8} ohm.m$

Length of wire, $L= 2.05 km$

Temperature, $T= 20^oC$

We know that relation between volume and density is given as

$Density= \frac{mass}{Volume}$

Therefore, volume occupied by one gram gold is given as,

$V=\frac{.001 kg}{19.3\times 10^3 Kg m^{-3}} = 5.181\times 10^{-8} m^3$ .........(1)

We Know that Volume of gold wire which is cylindrical in shape is given by following formula

$V=\pi \times r^2 \times L$ ......(2)

Here,

$A= \pi \times r^2$ ...........(3)

here A is the cross sectional area of cylendrical gold wire

From equation 2 and 3

we got

$V=A \times L$ ...............(4)

on comparing equation 1 and equation 4, we got,

$A \times L=5.181\times 10^{-8} m^3$

$A=\frac{5.181\times 10^{-8} m^3}{2050 m}$

$A=2.53\times 10^{-11}m^2$

we know that resistance and resistivity are related by following formula,

$Resistance = resistivity\times \frac{L}{A}$ ................(5)

Put values of resistivity, A and L in equation 5, we got

$R = \frac{2.44 \times 10^{-8} ohm.m \times 2050 m}{2.53\times 10^{-11} m^2}$

$R=1977 \times 10^3 ohm$

Therefore resistance of gold wire, $R=1977 \times 10^3 ohm$

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3 years ago

3 a There is a thin layer of water between the blade and the ice. Suggest how this affects friction .

Gelneren [198K]

Answer:

The water acts like a lubricant therefore has a smooth motion over the ice.

8 0

3 years ago