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Oksana_A [137]
3 years ago
14

List examples of foliated and non-foliated rocks. Explain the difference between the two types of metamorphic rocks.

Physics
1 answer:
igomit [66]3 years ago
4 0

Non-foliated metamorphic rocks are rocks that have been changed by heat and pressure into rocks with a non-layered or banded appearance. Some examples of non-foliated metamorphic rocks include quartzite, marble, amphibolite and hornfels.


Non-foliated metamorphic rocks are rocks that have been changed by heat and pressure into rocks with a non-layered or banded appearance. Some examples of non-foliated metamorphic rocks include quartzite, marble, amphibolite and hornfels.

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How much heat is absorbed by a 67g iron skillet when its temperature rises from 8C to 94C? ___ Joules
Katen [24]
I googled the specific heat capacity for iron so it might be off to the value you might have used, bit it should be close.

3 0
3 years ago
A temperature of a 0.349 kg sample of copper decreases from 87.0 °C to 21.0 °C. How much heat
Arte-miy333 [17]

We have that heat flow out of the copper sample  is mathematically given as

Q=8.86809J

<h3> Heat flow</h3>

Question Parameters

  • A temperature of a 0.349 kg sample of copper
  • decreases from 87.0 °C to 21.0 °C.

Endothermic Reaction

It is common knowledge that an endothermic reaction is one where the system absorbs heat energy from the <em>environs</em>.

Exothermic Reaction

An exothermic reaction is one where the system releases heat energy to the <em>environs</em>.

Generally the equation for Heat   is mathematically given as

Q=mCdt\\\\Therefore\\\\Q=0.349*0.385*(87-21)

Q=8.86809J

For more information on temperature visit

brainly.com/question/15267055

7 0
2 years ago
Which element(s) is/are not balanced in this equation?
Shkiper50 [21]
All elements are balanced. There are 1 Mg, 1 O, 2 Li's and 2 Cl's.
3 0
3 years ago
Consider a steel guitar string of initial length L=1.00 meter and cross-sectional area A=0.500 square millimeters. The Young's m
ira [324]

Answer:

\Delta L=15\,mm

Explanation:

Given:

  • length of a steel-string, L=1m
  • area of the string, A=0.5\,mm^2
  • Young's modulus of the steel, Y=2\times 10^{11} Pa
  • force of tension on the string, F=1500\,N

We have the relation for change in length:

\Delta L=\frac{F.L}{A.Y}

\Delta L=\frac{1500\times 1000}{0.5\times 10^{-6}\times 2\times 10^{11}}

\Delta L=0.015m

\Delta L=15\,mm

6 0
3 years ago
A 2.0 kg hanging mass stretches a coiled spring by 0.15 m. The spring constant, k, is: (A) 0.075 N/m, (B) 2.9 N/m (C) 131 N/m, (
Alla [95]

Answer:

C

Explanation:

Givens

m = 2 kg

F = 2 * 9.81

F =  19.62 N

x = 0.15 m

Formula

F = k*x

Solution

19.62 = k*0.15

k = 19.62/0.15

k = 130.8 which rounded to the nearest given answer is C

3 0
3 years ago
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