Answer:
C. The process through which distinct layers with characteristic chemical and/or physical properties are formed
Explanation:
I know It cause i took that test last year. Just wanna help GL.
Solution:
initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s
So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m
h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º
According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.
To solve this problem we must apply the concept related to the conservation of energy theorem.
PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so
PART B) Replacing the values given as,
Therefore the speed of the masses would be 1.8486m/s
Answer: vl = 2.75 m/s vt = 1.5 m/s
Explanation:
If we assume that no external forces act during the collision, total momentum must be conserved.
If both cars are identical and also the drivers have the same mass, we can write the following:
m (vi1 + vi2) = m (vf1 + vf2) (1)
The sum of the initial speeds must be equal to the sum of the final ones.
If we are told that kinetic energy must be conserved also, simplifying, we can write:
vi1² + vi2² = vf1² + vf2² (2)
The only condition that satisfies (1) and (2) simultaneously is the one in which both masses exchange speeds, so we can write:
vf1 = vi2 and vf2 = vi1
If we call v1 to the speed of the leading car, and v2 to the trailing one, we can finally put the following:
vf1 = 2.75 m/s vf2 = 1.5 m/s
Answer:
0.6 m
Explanation:
When a spring is compressed it stores potential energy. This energy is:
Ep = 1/2 * k * x^2
Being x the distance it compressed/stretched.
When the spring bounces the ice cube back it will transfer that energy to the cube, it will raise up the slope, reaching a high point where it will have a speed of zero and a potential energy equal to what the spring gave it.
The potential energy of the ice cube is:
Ep = m * g * h
This is vertical height and is related to the distance up the slope by:
sin(a) = h/d
h = sin(a) * d
Replacing:
Ep = m * g * sin(a) * d
Equating both potential energies:
1/2 * k * x^2 = m * g * sin(a) * d
d = (1/2 * k * x^2) / (m * g * sin(a))
d= (1/2 * 25 * 0.1^2) / (0.05 * 9.81 * sin(25)) = 0.6 m
