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serg [7]
2 years ago
5

The rotational kinetic energy term is often called the kinetic energy in the center of mass, while the translational kinetic ene

rgy term is called the kinetic energy of the center of mass. You found that the total kinetic energy is the sum of the kinetic energy in the center of mass plus the kinetic energy of the center of mass. A similar decomposition exists for angular and linear momentum. There are also related decompositions that work for systems of masses, not just rigid bodies like a dumbbell. It is important to understand the applicability of the formula Ktot=Kr+Kt. Which of the following conditions are necessary for the formula to be valid? Check all that apply.
a. The velocity vector V must be perpendicular to the axis of rotation.
b.The velocity vector must be perpendicular or parallel to the axis of rotation.
c.The moment of inertia must be taken about an axis through the center of mass.
Physics
2 answers:
weqwewe [10]2 years ago
6 0

Answer:

C

Explanation:

The total kinetic energy is the sum of the kinetic energy in the center of mass (Rotational Kinetic energy) plus the kinetic energy of the center of mass( Translational Kinetic Energy).

The formula

K_{tot} = K_{t} +K_{r}  is applicable only when

The moment of inertia must be taken about an axis through the center of mass.

Aneli [31]2 years ago
5 0

Answer:

c.The moment of inertia must be taken about an axis through the center of mass.

Explanation:

The movements of rigid bodies can always be divided into the translation movement of the center of mass and that of rotation around the center of mass. However, we can demonstrate that this is true for the kinetic energy of a rigid body that has both translational and rotational movement.

In this case the kinetic energy of the body is the sum of a part associated with the movement of the center of mass and another part associated with the rotation about an axis passing through the center of mass. This is all represented by the form Ktot = Kr + Kt, however we must consider that the moment of inertia must be taken around an axis through the center of mass, since rigid bodies at rest tend to remain at rest.

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The net force on a boat causes it to accelerate at 1.55 m/s2. The mass of the boat is 215 kg. The same net force causes another
jekas [21]

Answer:

2666 kg

0.11567 m/s²

Explanation:

m = Mass of boat

a = Acceleration of boat

From Newton's second law

Force

F=ma\\\Rightarrow F=215\times 1.55\\\Rightarrow F=333.25\ N

Force on the first boat is 333.25 N

F=ma\\\Rightarrow m=\frac{F}{a}\\\Rightarrow m=\frac{333.25}{0.125}\\\Rightarrow m=2666\ kg

Hence, mass of the second boat is 2666 kg

Combined mass = 2666+215 = 2881 kg

F=ma\\\Rightarrow a=\frac{F}{m}\\\Rightarrow a=\frac{333.25}{2881}\\\Rightarrow a=0.11567\ m/s^2

The acceleration on the combined mass is 0.11567 m/s²

6 0
2 years ago
Imagine that a small boy and his much larger father are enjoying a winter morning, sliding along the ice on a nearby playground.
kramer
You can write an hypothesis such as this:
The weight of an object has effects on the operating frictional force, the greater the weight, the higher the operating frictional force.
The father is the one with the higher weight while the son has the lower weight. The operating frictional force is the friction that their weights exert.
8 0
2 years ago
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A solid uniform sphere of mass 1.85 kg and diameter 45.0 cm spins about an axle through its center. Starting with an angular vel
KengaRu [80]

Answer:

The net torque is 0.0372 N m.

Explanation:

A rotational body with constant angular acceleration satisfies the kinematic equation:

\omega^{2}=\omega_{0}^{2}+2\alpha\Delta\theta (1)

with ω the final angular velocity, ωo the initial angular velocity, α the constant angular acceleration and Δθ the angular displacement (the revolutions the sphere does). To find the angular acceleration we solve (1) for α:

\frac{\omega^{2}-\omega_{0}^{2}}{2\Delta\theta}=\alpha

Because the sphere stops the final angular velocity is zero, it's important all quantities in the SI so 2.40 rev/s = 15.1 rad/s and 18.2 rev = 114.3 rad, then:

\alpha=-\frac{-(15.1)^{2}}{2(114.3)}=1.00\frac{rad}{s^{2}}

The negative sign indicates the sphere is slowing down as we expected.

Now with the angular acceleration we can use Newton's second law:

\sum\overrightarrow{\tau}=I\overrightarrow{\alpha} (2)

with ∑τ the net torque and I the moment of inertia of the sphere, for a sphere that rotates about an axle through its center its moment of inertia is:

I = \frac{2MR^{2}}{5}

With M the mass of the sphere an R its radius, then:

I = \frac{2(1.85)(\frac{0.45}{2})^{2}}{5}=0.037 kg*m^2

Then (2) is:

\sum\overrightarrow{\tau}=0.037(-1.00)=0.037 Nm

7 0
2 years ago
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Jack's model includes one object consisting of two molecules and one object consisting of six molecules. He put 16 energy cubes
Novay_Z [31]

Answer:

the molecules have different energy and the system is not in equilibrium

Explanation:

The model developed by Jack has the same energy for each of the two objects, but as each object is made up of a different number of molecules, in the system with more molecules, object 2, each one has approximately 2.4 less energy and the molecules of Object 1 have an E / 2 energy.

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5 0
2 years ago
A 5 kg block moves with a constant speed of 10 ms to the right on a smooth surface where frictional forces are considered to be
nirvana33 [79]

Answer:

Work done, W = 19.6 J

Explanation:

It is given that,

Mass of the block, m = 5 kg

Speed of the block, v = 10 m/s

The coefficient of kinetic friction between the block and the rough section is 0.2

Distance covered by the block, d = 2 m

As the block passes through the rough part, some of the energy gets lost and this energy is equal to the work done by the kinetic energy.

W=\mu_kmgd

W=0.2\times 5\times 9.8\times 2

W = 19.6 J

So, the change in the kinetic energy of the block as it passes through the rough section is 19.6 J. Hence, this is the required solution.

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2 years ago
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