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nika2105 [10]
3 years ago
11

Why do intrusive igneous rocks have large crystals?

Physics
1 answer:
Aleksandr-060686 [28]3 years ago
3 0
The magma cools very slowly so that the crystals of different minerals have time to grow. The magma molecules in have time to organize themselves into crystal formations before the magma hardens. The naked eye helps the Intrusive rocks that has large crystals be seen.
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There are correlation and causation between the force of the finger and the movement of the books
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In the reaction 2Ca + O2 → 2CaO for every 2 Ca you will need how much O2?
Serjik [45]

Answer:

1

Explanation: that is the ratio

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2 years ago
An object with a higher temperature can have less thermal energy than an object
Anna [14]
It’s true, because it also depends on things like mass. Higher temperature but less mass< Lower temperature but more mass.
4 0
2 years ago
What is the difference between a negative feedback system and a positive feedback system?
Anni [7]

Answer:

D

Explanation:

The negative feedback is responsible for maintaining equilibrium (stability) in a system as it lessens effects, which is contrary to positive feedback which increases reaction and moves a system further away from equilibrium (stability), It does so by amplifying the effects of a product or event and occurs when something needs to happen quickly. e.g

  • Insulin lowers down blood sugar levels, so when the body detects that it has too much sugar, the pancreas is prompted to release insulin and only stops when balance is achieved; hence, negative feedback.  
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7 0
3 years ago
Water flows through a cast steel pipe (k = 50 W m.K, ε = 0.8) with an outer diameter of 104mm and 2 mm wall thickness. Calculate
masha68 [24]

Answer:

The heat loss per unit length is   \frac{Q}{L}   = 2981 W/m

Explanation:

From the question we are told that

     The outer diameter of the pipe is d = 104mm = \frac{104}{1000} = 0.104 m

     The thickness is  D = 2mm = \frac{2}{1000} = 0.002m  

      The temperature  of water is  T = 90^oC = 90 + 273 = 363K  

      The outside air temperature is T_a = -10^oC = -10 +273 = 263K

        The water side heat transfer coefficient is z_1 = 300 W/ m^2 \cdot K

       The  heat transfer coefficient is  z_2 = 20 W/m^2 \cdot K

The heat lost per unit length is mathematically represented as

           \frac{Q}{L}   = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1}  +  \frac{ln [\frac{d}{D} ]}{z_2}}

Substituting values

         \frac{Q}{L}   = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300}  +  \frac{ln [\frac{0.104}{0.002} ]}{20}}

           \frac{Q}{L}   = \frac{628}{0.2107}

           \frac{Q}{L}   = 2981 W/m

6 0
3 years ago
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