Answer:
θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²
Explanation:
This is an angular kinematic exercise the equation for the angular position
the particle A
θ = θ₀ + ω₀ t + ½ α t²
They say for the particle B
w₀B = ½ w₀
αB = 2 α
In addition, the particle begins at a time t_1 after particle A, in order to use the same timer, we must subtract this time from the initial
t´ = t - t_1
l
et's write the equation of particle B
θ = θ₀ + w₀B t´ + ½ αB t´2
replace
θ = θ₀ + ½ w₀ (t -t_1) + ½ 2α (t -t_1)²
θ = θ₀ + ½ w₀ (t -t_1) + α (t -t_1)²
<span>For hydrolysis to monosaccharides, one molecule of a disaccharide needs only one molecule of water.
C12H22O11 (sucrose) + H2O = C6H12O6 (glucose) + C6H12O6 (fructose)
Structurally, a disaccharide molecule may be viewed as a product formed by the condensation of two molecules of monosaccharides with the elimination of a water molecule. So, only one H2O molecule is needed for the reverse process.</span>
It took so long because at the time there was no way for people to study the behavior formally. im not sure what helped it get recognized but i know wihelm wundt helped get it recongnized.
sorry i couldnt be much help
Answer:
no:
Explanation:
it would grow and no longer be able to fit through the loop due to the hot air expanding.
Answer:
The speed of the ball is 42.5 m/s
Explanation:
The initial kinetic energy of the ball is:
= 85.75 J
The speed of the ball after leaving the bat is:
V=47.92 m/s
Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:
