C. Temperature, chemical composition and mineral structure
Explanation:
The Bowen's reaction series illustrates the relationship between temperature, chemical composition and mineral structure.
The series is made up of a continuous and discontinuous end through which magmatic composition can be understood as temperature changes.
- The left part is the discontinuous end while the right side is the continuous series.
- From the series, we understand that a magmatic body becomes felsic as it begins to cool to lower temperature.
- A magma at high temperature is ultramafic and very rich in ferro-magnesian silicates which are the chief mineral composition of olivine and pyroxene. These minerals are predominantly found in mafic- ultramafic rocks. Also, we expect to find the calcic-plagioclase at high temperatures partitioned in the magma.
- At a relatively low temperature, minerals with frame work structures begins to form . The magma is more enriched with felsic minerals and late stage crystallization occurs here.
Learn more:
Silicate minerals brainly.com/question/4772323
#learnwithBrainly
Answer:Chemical
Explanation:Photosynthesis is the biochemical reaction through which all green plants, including trees, make their food to live and grow.
Answer:
speed of puck acc. to the radar gun = 138 km/h
speed of player = 15 km/h
since the player is in motion when he shoots, the speed of the puck will be the sum of the speed of the player and the speed at which he shot. so,
speed of puck = speed of player + speed of puck acc. to player
138 = 15 + speed of puck acc. to player
speed of puck acc. to player = 138 -15
speed of puck acc. to player = 123 km/h
Brainly this answer if you think it deserves it
Answer:
weight at height = 100 N .
Explanation:
The problem relates to variation of weight due to change in height .
Let g₀ and g₁ be acceleration due to gravity , m is mass of the object .
At the surface :
Applying Newton's law of gravitation
mg₀ = G Mm / R²
At height h from centre
mg₁ = G Mm /h²
Given mg₀ = 400 N
400 = G Mm / R²
400 = G Mm / (6400 x 10³ )²
G Mm = 400 x (6400 x 10³ )²
At height h from centre
mg₁ = 400 x (6400 x 10³ )²/ ( 2 x 6400 x 10³)²
= 400 / 4
= 100 N .
weight at height = 100 N
Answer:
Aluminum
Explanation:
promise
give me brainlest plleaseee