Answer:
The upper limit of metamorphism occurs at the pressure and temperature of wet partial melting of the rock in question. Once melting begins, the process changes to an igneous process rather than a metamorphic process. During metamorphism the protolith undergoes changes in texture of the rock and the mineral make up of the rock.
Explanation:
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Answer:
1 P = 0.5
2 P = 0.3
3 P = 0.01
Explanation:
The probability formula is
![P =V^N](https://tex.z-dn.net/?f=P%20%3DV%5EN)
Where P is the probability V is the volume while N is the number of distinguishing particles
So for N = 1 and ![V = 0.500m^3](https://tex.z-dn.net/?f=V%20%3D%200.500m%5E3)
![P = (0.500m^3)^1\\](https://tex.z-dn.net/?f=P%20%3D%20%280.500m%5E3%29%5E1%5C%5C)
= 0.5
For N = 1 and ![V = 0.300m^3](https://tex.z-dn.net/?f=V%20%3D%20%200.300m%5E3)
![P = (0.300m^3)^1](https://tex.z-dn.net/?f=P%20%3D%20%280.300m%5E3%29%5E1)
= 0.3
For N = 1 and ![V = 0.0100m^3](https://tex.z-dn.net/?f=V%20%3D%200.0100m%5E3)
![P =(0.0100m^3)^1](https://tex.z-dn.net/?f=P%20%3D%280.0100m%5E3%29%5E1)
= 0.01
Answer:
Option B. 5 nC
Explanation:
From the question given above, the following data were obtained:
Capicitance (C) = 100 pF
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:
1 pF = 1×10¯¹² F
Therefore,
100 pF = 100 pF × 1×10¯¹² F / 1 pF
100 pF = 1×10¯¹⁰ F
Next, we shall determine the quantity of charge. This can be obtained as follow:
Capicitance (C) = 1×10¯¹⁰ F
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Q = CV
Q = 1×10¯¹⁰ × 50
Q = 5×10¯⁹ C
Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:
1 C = 1×10⁹ nC
Therefore,
5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C
5×10¯⁹ C = 5 nC
Thus, the quantity of charge is 5 nC
Answer:
F=ma
Explanation:
Force = mass * acceleration
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