Multiple-Concept Example 6 reveiws the principles that play a role in this problem. A nuclear power reactor generates 2.3 x 109
2 answers:
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The characteristics of the kinematics allow to find the results for the questions about the movement of the body are:
a) we have four sections;
- 0 to 8 s The body is accelerating.
- 8 to 10 s The body goes at a constant speed, the acceleration is zero.
- 10 to 14 Body accelerating.
- 14 to 18 Body slowing down.
b) The acceleration is the first 8 s is: a = 0.25 m / s²
c) The maximum acceleration is: a = 1 m / s²
d) The displacement is: i) d₁ = 8m, ii) = 16 m
e) maximum speed is: v = 6 m / s
Kinematics studies the movement of bodies by finding relationships between the position, speed and acceleration of bodies.
v = v₀ + a t
y = v₀ t + ½ a t²
where v and v₀ is the current and initial velocity, respectively, a is the acceleration and t is time.
In many circumstances graphs are made for their analysis, in a graph of speed versus time when we have a horizontal line the speed is constant, the acceleration is zero and in the case of a slope there is an acceleration, we have two cases:
- Positive slope the body is accelerating and the speed is increasing.
- Negative slope the body is stopping, the speed decreases.
Let's answer the different questions about the system.
a) in the attached we have a graph of the velocity versus time, each section corresponds to a change in the slope of the graph, we have four sections;
- 0 to 8 s The body is accelerating.
- 8 to 10 s The body goes at a constant speed, the acceleration is zero.
- 10 to 14 Body accelerating.
- 14 to 18 Body slowing down.
b) The acceleration is the first 8 s
v = v₀ + a t
a = 0.25 m / s²
c) The maximum acceleration is when the slope is maximum.
a = 1 m / s²
Therefore the acceleration is maximum in the section between 10 and 14 s
d) The total displacement is the sum of the displacements of each section.
We look for every displacement.
d₁ = v₀ + ½ a₁ Δt²
d₁ = 0 + ½ 0.25 8²
d₁ = 8 m
In the second section the velocity is constant
d₂ = v₂ Δt₂
d₂ = 2 (10-8)
d₂ = 4 m
The third section.
d₃ = v₀ + ½ a t²
d₃ = 2 + ½ 1 (14-10) ²
d₃ = 10 m
The distance of the fourth section.
we look for acceleration
a₄ =
a₄ =
a₄ = -1.5 m / s²
d₄ = 6 + ½ (-1.5) (1814) ²
d₄ = -6 m
The total displacement is;
= 8 + 4 + 10 -6
= 16 m
e) The maximum speed is the highest point in the graph of speed versus time that in the attachment we can see corresponds to
v = 6 m / s
In conclusion using the characteristics of kinematics we can find the results for the questions about the motion of bodies are:
a) we have four sections;
- 0 to 8 s The body is accelerating.
- 8 to 10 s The body goes at a constant speed, the acceleration is zero.
- 10 to 14 Body accelerating.
- 14 to 18 Body slowing down.
b) The acceleration is the first 8 s is: a = 0.25 m / s²
c) The maximum acceleration is: a = 1 m / s²
d) The displacement is: i) d₁ = 8m, ii) = 16 m
e) maximum speed is: v = 6 m / s
Learn more about kinematics here: brainly.com/question/24783036
Answer:
K.E = 0.0075 J
Explanation:
Given data:
Mass of the ball = 1.5 kg
radius, r = 50 cm = 0.5 m
Angular speed, ω = 12 rev/min = (12/60) rev/sec = 0.2 rev/sec
Now,
the kinetic energy is given as:
K.E =
where,
I is the moment of inertia = mr²
on substituting the values, we get
or
K.E = 0.0075 J