Explanation:
If there is no no pictures how I will describe???????????
The vertical movement of the projectile is described by:
y = H - gt² / 2
When the cannonball is on the ground, y= 0 so:
0 = H - gt² /2
Solving for t:
The horizontally movement of the projectile is described by:
x = v₀t
Solving for v₀:
v₀ = x/t
v₀ = 175 m / 2.8 s
v₀ = 61.5 m/s
The light reactions could be viewed as analogous to a hydroelectric dam. In that case, the wall of the dam that holds back the water would be analogous to the thylakoid membrane.
Thylakoid membrane is present in cyanobacteria and chloroplasts of plants. It plays a crucial role in photosynthesis and photosystem II reactions.
In general, these are the regions where light-dependent reactions take place. The thylakoid membrane is a lipid-bound membrane that maintains potential difference and also controls the flow of liquids across the membrane during light reactions.
In the provided case, we can see that the wall of the dam holds back the water, similarly, in light-dependent reactions, thylakoid membranes control the liquid flow and also regulate the potential gradient across the membrane and also allow the selective proteins to pass through.
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Answer:
The total surface are of the bowl is given by: 0.0532*pi m² (approximately 0.166533 m²)
Explanation:
The total surface area of the semi-spherical bowl can be decomposed in three different sections: 1) an outer semi-sphere of radius 12 cm, 2) an inner semi-sphere of radius 10 cm, and 3) the edge, which is a 2-dimensional ring with internal radius of 10 cm and external radius of 12 cm. We will compute the areas independently and then sum them all.
a) Outer semi-sphere:
A1 = 2*pi*r² = 2*pi*(12 cm)² = 288*pi cm² = 904.78 cm²
b) Inner semi-sphere:
A2 = 2*pi*(10 cm)² = 200*pi cm² = 628.32 cm²
c) Edge (Ring):
A3 = pi*(r1² - r2²) = pi*((12 cm)²-(10 cm)²) = pi*(144-100) cm² = 44*pi cm² = 138.23 cm²
Therefore, the total surface area of the bowl is given by:
A = A1 + A2 + A3 = 288*pi cm² + 200*pi cm² + 44*pi cm² = 532*pi cm² (approximately 1665.33 cm²)
Changing units to m², as required in the problem, we get:
A = 532*pi cm² * (1 m² / 10, 000 cm²) = 0.0532*pi m² (approximately 0.166533 m²)
Answer:
yes it flows through flow paths.
Explanation:
