Stark contrast to paths on energy surfaces or even mechanistic reactions, rule-based and inductive computational approaches to reaction prediction mostly consider only overall transformations. Overall transformations are general molecular graph rearrangements reflecting only the net change of several successive mechanistic reactions. For example, Figure 1 shows the overall transformation of an alkene interacting with hydrobromic acid to yield the alkyl bromide along with the two elementary reactions which compose the transformation.
Answer:
Explanation:
Use the one-dimensional equation
where vf is the final velocity of the dog, v0 is the initial velocity of the dog, a is the acceleration of the dog, and t is the time it takesto reach that final velocity. For us:
0 = 2 + -.43t and
-2 = -.43t so
t = 4.7 seconds
Answer:
a) 1.73*10^5 J
b) 3645 N
Explanation:
106 km/h = 106 * 1000/3600 = 29.4 m/s
If KE = PE, then
mgh = 1/2mv²
gh = 1/2v²
h = v²/2g
h = 29.4² / 2 * 9.81
h = 864.36 / 19.62
h = 44.06 m
Loss of energy = mgΔh
E = 780 * 9.81 * (44.06 - 21.5)
E = 7651.8 * 22.56
E = 172624.6 J
Thus, the amount if energy lost is 1.73*10^5 J
Work done = Force * distance
Force = work done / distance
Force = 172624.6 / (21.5/sin27°)
Force = 172624.6 / 47.36
Force = 3645 N
Answer:
TRUE - In any collision between two objects, the colliding objects exert equal and opposite force upon each other. This is simply Newton's law of action-reaction.
Answer:
(a) the speed of the block after the bullet embeds itself in the block is 3.226 m/s
(b) the kinetic energy of the bullet plus the block before the collision is 500J
(c) the kinetic energy of the bullet plus the block after the collision is 16.13J
Explanation:
Given;
mass of bullet, m₁ = 0.1 kg
initial speed of bullet, u₁ = 100 m/s
mass of block, m₂ = 3 kg
initial speed of block, u₂ = 0
Part (A)
Applying the principle of conservation linear momentum, for inelastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
v is the speed of the block after the bullet embeds itself in the block
(0.1 x 100) + (3 x 0) = v (0.1 + 3)
10 = 3.1v
v = 10/3.1
v = 3.226 m/s
Part (B)
Initial Kinetic energy
Ki = ¹/₂m₁u₁² + ¹/₂m₂u₂²
Ki = ¹/₂(0.1 x 100²) + ¹/₂(3 x 0²)
Ki = 500 + 0
Ki = 500 J
Part (C)
Final kinetic energy
Kf = ¹/₂m₁v² + ¹/₂m₂v²
Kf = ¹/₂v²(m₁ + m₂)
Kf = ¹/₂ x 3.226²(0.1 + 3)
Kf = ¹/₂ x 3.226²(3.1)
Kf = 16.13 J
