Chanel has some cotton candy that came in a cloudy shape. She wants to make it more dense. Which describes the candy before and
2 answers:
Answer: Option (C) is the correct answer.
Explanation:
As we known that density is the amount of mass divided by volume of the substance.
Mathematically, Density =
So, when candy was present in the shape of a cloud then it means that it was fluffy as it has more volume.
Since, density is inversely proportional to volume therefore, with increase in volume there will occur a decrease in density.
But when the candy will become compact then there will occur a decrease in its volume. Hence, then there will occur an increase in the density of the candy.
Thus, we can conclude that the statement candy before was fluffy, and the candy after was compacted best describes the candy before and after Chanel manipulated it.
You might be interested in
With constant angular acceleration , the disk achieves an angular velocity
at time
according to
and angular displacement according to
a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of
b. Under constant acceleration, the average angular velocity is equivalent to
where and
are the final and initial angular velocities, respectively. Then
c. After 1.00 s, the disk has instantaneous angular velocity
d. During the next 1.00 s, the disk will start moving with the angular velocity equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle
according to
which would be equal to
Answer:
a) = 0.25 m / s b) u = 0.25 m / s
Explanation:
a) To solve this problem let's start with the conservation of the moment, for this we define a system formed by the ball plus the dog, in this case all the forces are internal and the moment is conserved
We will write the data
m₁ = 0.40 kg
v₁₀ = 9.0 m / s
m₂ = 14 kg
v₂₀ = 0
Initial
po = m₁ v₁₀
Final
= (m₁ + m₂) vf
po = pf
m₁ v₁₀ = (m₁ + m₂)
= v₁₀ m₁ / (m₁ + m₂)
= 9.0 (0.40 / (0.40 +14)
= 0.25 m / s
b) This is the reference frame of the center of mass of the system in this case the speed of this frame is the speed of the center of mass
u = 0.25 m / s
In the direction of movement of the ball
c) Let's calculate the kinetic energy in both moments
Initial
K₀ = ½ m₁ v₁₀² +0
K₀ = ½ 0.40 9 2
K₀ = 16.2 J
Final
= ½ (m₁ + m₂)
2
= ½ (0.4 +14) 0.25 2
= 0.45 J
ΔK = K₀ -
ΔK = 16.2-0.445
ΔK = 1575 J
These will transform internal system energy
d) In order to find the kinetic energy, we must first find the velocities of the individual in this reference system.
v₁₀’= v₁₀ -u
v₁₀’= 9 -.025
v₁₀‘= 8.75 m / s
v₂₀ ‘= v₂₀ -u
v₂₀‘= - 0.25 m / s
‘=
- u
= 0
Initial
K₀ = ½ m₁ v₁₀‘² + ½ m₂ v₂₀‘²
Ko = ½ 0.4 8.75² + ½ 14.0 0.25²
Ko = 15.31 + 0.4375
K o = 15.75 J
Final
= ½ (m₁ + m₂) vf’²
= 0
All initial kinetic energy is transformed into internal energy in this reference system