Answer:
it would be c i just had it
Explanation:
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<span>Trichome density and type and cannabinoid content of leaves and bracts were quantitated during organ ontogeny for three clones of Cannabis sativa L. Trichome initiation and development were found to occur throughout leaf and bract ontogeny. On leaves, bulbous glands were more abundant than capitate-sessile glands for all clones, although differences in density for each gland type were evident between clones. On pistillate bracts, capitate-sessile glands were more abundant than the bulbous form on all clones, and both types decreased in relative density during bract ontogeny for each clone. The capitate-stalked gland, present on bracts but absent from vegetative leaves, increased in density during bract ontogeny. The capitate-stalked gland appeared to be initiated later than bulbous or capitate-sessile glands during bract development and on one clone it was first found midway in bract ontogeny. Nonglandular trichomes decreased in density during organ ontogeny, but the densities differed between leaves and bracts and also between clones. Specific regulatory mechanisms appear to exist to control the development of each trichome type independently.</span>
We will apply the concepts related to Newton's second law. At the same time we will convert everything to the system of international units.

The values of the velocities are,


We know that the acceleration is equivalent to the change of the speed in a certain time therefore



Now applying the Newton's second law we have,



Therefore the approximate magnitude is 8516.36N
The problem is solved and the questions are answered below.
Explanation:
a. To calculate the speed of the 0.66 kg ball just before the collision
V₀ + K₀ = V₁ + K₁
= mgh₀ = 1/2 mv₁²
where, h= r - r cosθ
V = 
V = 2.42 m/s
b. Calculate the speed of the 0.22 kg ball immediately after the collision
y = y₀ + Vy₀t - 1/2 gt²
0 = 1.2 - 1/2 gt²
t = 0.495 s
x = x₀ + Vx₀t
1.4 = 0 + vx₀ (0.495)
Vx₀ = 2.83 m/s
C. To Calculate the speed of the 0.66 kg ball immediately after the collision
m₁ v₁ = m₁ v₃ + m₂ v₄
(0.66)(2.42) = (0.66) v₃ + (0.22)(2.83)
V₃ = 1.48 m/s
D. To Indicate the direction of motion of the 0.66 kg ball immediately after the collision is to the right.
E. To Calculate the height to which the 0.66 kg ball rises after the collision
V₀ + k₀ = V₁ + k₁
1/2 mv₀² = mgh₁
h₁ = v₀²/2 g
= 0.112 m
F. Based on your data, No the collision is not elastic.
Δk = 1/2 m₁v₃² =1/2 m₂v₄² - 1/2 m₁v₁²
= 1/2 (0.66)(1.48)² + 1/2 (0.22)(2.83)² - 1/2 (0.66)(2.42)²
= - 0.329 J
Hence, kinetic energy is not conserved.