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Brilliant_brown [7]
3 years ago
9

Describe the patterns created by minerals in foliated rock.

Physics
1 answer:
professor190 [17]3 years ago
3 0
Foliated Metamorphic Rocks. Foliated metamorphic rocks are formed within the Earth's interior under extremely high pressures that are unequal, occurring when the pressure is greater in one direction than in the others (directed pressure). ... This reduces the overall pressure on the rock and gives it a stripped look.
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Us your understanding of asexual reproduction to explain why is it important that organisms reproduce in a variety of ways.
valentina_108 [34]
The genetic material is identical in asexual reproduction- in order for organisms to be strong they need variety so if a disease comes, some of the species may be able to fight it off because of their varied genetics
5 0
3 years ago
A concave mirror is used as a<br> reflector ina torch light?
4vir4ik [10]

Answer:

A concave mirror is used as a torch reflector. ... When a light bulb is placed at the focus of a concave mirror reflector, the diverging light rays of the bulb are collected by the reflector. These rays are then reflected to produce a strong, parallel-sided beam of light.

Explanation:

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Shorter the vibrating part more will be the pitch. How?​
Lina20 [59]

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6 0
2 years ago
Two elements are found in the same column of the periodic table. You would expect these elements to have
konstantin123 [22]
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8 0
3 years ago
Read 2 more answers
Cierto volumen de gas se encuentra a 60°c de temperatura y 5atm de presión, es calentado hasta 140°c, estado en el cual ocupa un
earnstyle [38]

Answer:

V1 = 2221.33 L

Explanation:

The system is about a ideal gas. Then you can use the equation for ideal gases for a volume V1, temperature T1 and pressure P1:

P_1V_1=nRT_1   (1)

And also for the situation in which the variables T, V and P has changed:

P_2V_2=nRT_2   (1)

R: constant of ideal gases = 0.082 L.atm/mol.K

For both cases (1) and (2) the number of moles are the same. Next, you solve for n in (1) and (2):

n=\frac{P_1V_1}{RT_1}\\\\n=\frac{P_2V_2}{RT_2}

Next, you equal these equations an solve for T2:

\frac{P_1V_1}{RT_1}=\frac{P_2V_2}{RT_2}\\\\V_1=\frac{P_2V_2T_1}{P_1T_2}

Finally you replace the values of P2, V2, T1 and T2:

V_1=\frac{(7atm)(680L)(140\°C)}{(60\°C)(5atm)}=2221.33\ L

Hence, the initial volume of the gas is 2221.33 L

5 0
3 years ago
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