Acceleration is rate of change of_

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

2 years ago

A 2.0 kg particle moving along the z-axis experiences the

DochEvi [55]

At point x = 0, the particle accelerates. Since there will be change of velocity at that point. The the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.

Given that a 2.0 kg particle moving along the z-axis experiences the force shown in a given figure.

Force is the product of mass and acceleration. While acceleration is the rate of change of velocity. Both the force and acceleration are vector quantities. They have both magnitude and direction.

If the particle's velocity is 3.0 m/s at x = 0 m, that mean that the particle experience change of velocity at point x = 0. Since the the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.

Learn more here: brainly.com/question/20366032

6 0

2 years ago

This chart lists four examples of two objects that are in contact.

Oxana [17]

Before coming into conclusion first we have to understand the direction of heat flow.

Heat is the transferred thermal energy from one body to another body due to the temperature difference just like water flows from higher level to lower level.

Whenever two bodies having different temperature come closer to each other heat will flow from hotter body to cooler one if no external work is done. The heat flow may be through any of the ways i.e conduction,radiation or convection. Hence temperature difference is the parameter which gives the direction of heat flow.

The temperature is also considered as a measure of average kinetic energy of the substance.The thermal energy does not give the direction heat flow. Heat may flow from the body having low thermal energy but at higher temperature to the body having higher thermal energy but at low temperature. The reverse does not happen naturally .

In example 1 there is fire and air. Obviously fire is at high temperature and air at low temperature.So heat will flow from object 1 to object 2.

In example 2 there is a metal at 80 degree Celsius and another metal at 12 degree Celsius .So heat will flow from object 1 to object 2

In example 3 we have cooler ocean and warm air. So the heat will flow from object 2 to object 1.

In example 4 we have a tool with high thermal energy and a material with little thermal energy.We already know that thermal energy can not determine the direction of heat flow. Here the temperature of each substance is not given.The kinetic energy is part of thermal energy.So there is the chance of higher kinetic energy of the tool for having higher thermal energy .At that time the heat will flow object 1 to object 2.Otherwise the reverse will occur. So it is a special case.

As per the question only option 4 is correct which tells that heat will flow from object 1 to object 2 in examples 1,2,4, and heat will flow from object 2 to 1 in example 3. Other options violate the fundamental law of thermodynamics.

7 0

2 years ago

at the moment a truck traveling with a constant velocity of 13.6 m/s passes a dog at rest the dog begins moving with a constant

goblinko [34]

I don’t know how to help you hope you figure it out have a good day

4 0

1 year ago

When does air become turbulent around a thrown ball?

kherson [118]

Answer: B (on the rough side)

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8 0

2 years ago

Acceleration is rate of change of__?