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serg [7]
3 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]3 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]3 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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katrin [286]

Answer:

Phosphorus

Explanation:

Phosphorus needs 3 electrons to complete its outter most shell of electrons

8 0
3 years ago
Explain how the specific heat of water helps keep our bodies from overheating
choli [55]
This high specific heat of water prevents abrupt changes in temperature and helps the body maintain temperatures that are relatively constant.
5 0
3 years ago
Graphing: Michelle climbs a tree and drops the toy car once she has reached the top. Please create an Energy-Time graph to show
Vlada [557]

Answer:

Refer to the attachment for the graph. The shape of both functions should resemble part of a parabola. Assumption: air resistance on the car is negligible.

Explanation:

  1. The toy car started with a large amount of (gravitational) potential energy (PE) when it is at the top of the tree. Since it wasn't moving (as it was within Michelle's grip,) its kinetic energy (KE) would be equal to zero.
  2. As the car falls to the ground, its PE converts to KE.
  3. When the car was about to reach the ground, its PE is almost zero, while its KE is at its maximum.

<h3>PE of the car over time</h3>

The size of gravitational PE depends on both the mass and the height of the object. In this case, assume that the mass of the car stayed the same, PE should be proportional to the height of the car.

Assume that air resistance on the car is negligible. The height h of the car at time t could be found with the equation:

\displaystyle h = -\frac{1}{2}\, g\, t^2 + h_0 \, (\text{Initial height}),

where

  • g \approx \rm 9.81\; m \cdot s^{-2} near the surface of the earth, and
  • h_0 is the initial height of the car.

On the other hand, \displaystyle \text{GPE} = m \, g \, h = -\frac{m \cdot g^2}{2}\, t^2 + \underbrace{m \cdot g \cdot h_0}_{\text{Initial GPE}}.

In other words, plotting the gravitational PE of the car against time would give a parabola. Since \displaystyle -\frac{m \cdot g^2}{2} < 0 (the quadratic coefficient is smaller than zero,) the parabola should open downwards. Besides, since at t = 0 the initial GPE is positive, the y-intercept of this parabola should also be positive.

<h3>KE of the car over time</h3>

Assume that the air resistance on the car is negligible. The mechanical energy (ME) of the toy car should conserve (stay the same.) The mechanical energy of an object is the sum of its PE and KE. The PE of the toy car has already been found as a function of time. Therefore, simply subtract the expression of PE from mechanical energy to find an expression for KE.

To find the value of mechanical energy, consider the PE of the toy car before it was dropped. Since initially KE was equal to zero, the mechanical energy of the toy car would be equal to its initial PE. That's m \cdot g \cdot h_0. If there's no air resistance, the value of ME would stay at

Subtract PE from ME to obtain an expression for KE:

\begin{aligned} \text{KE} &= \text{ME} - \text{PE} \cr &= m \cdot g \cdot h_0 - \left(-\frac{m \cdot g^2}{2}\, t^2 + m \cdot g \cdot h_0\right)\cr &= \frac{m \cdot g^2}{2}\, t^2\end{aligned}.

That's also a parabola when plotted against t. Note that since the quadratic coefficient \displaystyle \frac{m \cdot g^2}{2} is positive, the parabola shall open upwards.

7 0
3 years ago
Which of the following is a balanced equation?
EastWind [94]
Please fell free to ask if you have any more confusion!

7 0
3 years ago
Suppose the current in a conductor decreases exponentially with time according to the equation I(t) = I₀ e**(-t / T) , where I₀
Anika [276]
  1. The quantity of charge that passes the point between t = 0 and t = τ is 0.6321I₀τ.
  2. The quantity of charge that passes the point between t = 0 and t = 10τ is 0.99995I₀τ.
  3. The quantity of charge that passes the point between t = 0 and t = ∞ is I₀τ.

<h3>How to determine the charge?</h3>

In an electric field, Euler's number suggest that current starts with a maximum value and decreases to 36.9% of the preceding numerical value in every time interval, τ.

Also, the current in this conductor decreases exponentially with time and as such, it would be used to derive an equation for charge through integration:

I = dQ/dt

dQ = Idt

By integrating, we have this general integral:

Q = ∫dQ = ∫Idt

At an end-time, T, the integral becomes:

Q = \int\limits^T_0 {I_0e^{\frac{-t}{\tau} } } \, dt

From time, t = 0 to time, t = T, we would integrate to have:

Q = \int\limits^T_0 {-I_0\tau)e^{\frac{-t}{\tau} } } \, (\frac{dt}{\tau} )

At t = 0, we set Q = 0 and obtain:

Q = {-I_0\tau(e^{\frac{-t}{\tau} } - e^0)} \, (\frac{dt}{\tau} )\\\\Q = I_0\tau(1-e^{\frac{-t}{\tau} })

Between t = 0 and t = τ, the charge that passes through this point is given by:

Q = I₀τ(1 - e⁻¹)

Q = 0.6321I₀τ.

Between t = 0 and t = 10τ, the charge that passes through this point is given by:

Q = I₀τ(1 - e⁻¹⁰)

Q = 0.99995I₀τ.

Between t = 0 and t = ∞, the charge that passes through this point is given by:

Q = I₀τ(1 - e⁻∞)

Q = I₀τ.

In conclusion, we can reasonably infer and logically deduce that the amount of current decreases sharply with respect to time and this makes the total charge it transports is finite.

Read more on current and charge here: brainly.com/question/15693063

#SPJ4

Complete Question:

Suppose the current in a conductor decreases exponentially with time according to the equation I(t) = I₀e^(-t / T) , where I₀ is the initial current (at. t = 0 ) and T is a constant. having dimensions of time. Consider a fixed observation point within the conductor.

(a) How much charge passes this point between t = 0 and t = τ?

(b) How much charge passes this point between t = 0 and t = 10τ?

(c) How much charge passes this point between t = 0 and t = ∞?

6 0
2 years ago
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