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Volgvan
4 years ago
8

And metamorphic rock is classified according to its

Physics
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.
3 0
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:

the univerce is VERY large so a, if im wrong i apologise :(

6 0
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

7 0
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
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