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4 years ago

Define power and describe how to determine power. (4 points)

1 answer:

7 0

Answer: The power is a measure of the rate at which work is done (or similarly, at which energy is transferred). The standard unit of power is Watt (1W). The power is determined by the change in energy, delta E (number of joules) and delta t, which is the time taken in seconds then: P= delta E/ delta t.

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Explanation:

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While standing in the middle of a mountain range, a camper is curious to see what would happen if he yelled loudly into the dist

Vera_Pavlovna [14]

Answer:

Reverberation

Explanation:

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4 years ago

Horizontal angulation is: Select one: a. the side-to-side angulation. b. different when using the paralleling and bisecting tech

statuscvo [17]

Horizontal angulation is Select one: a. the side-to-side angulation. b. different when using the paralleling and bisecting techniques.

Angles that are horizontal. relates to the central ray's placement in a horizontal, or side-to-side, plane.

A picture with an overlap of nearby structures in the horizontal plane is produced by situating the central ray such that the horizontal angulation is not directed through the interproximal contacts of the adjacent teeth (the contact areas of the teeth are superimposed over each other).

Fracture angulation refers to a particular kind of fracture displacement in which the bone's natural axis has been changed so that the distal end now points off in a different direction. When using a bite-wing tab, the x-ray beam's center ray must be pointed at the contact points between teeth. The x-ray beam must be centered on the receptor to guarantee that the receptor is exposed.

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8 0

2 years ago

How do you think television programs movies and car commercials influence driving behavior

AnnZ [28]

When you watch the same thing over and over (like a car commercial), then your brain begins to question morals. Like if you see a racing movie you might want to go and race. Or if you see an underage kid illegally driving, you might want to try it too.

4 0

3 years ago

An infinite line of charge with linear density λ1 = 8.2 μC/m is positioned along the axis of a thick insulating shell of inner r

bixtya [17]

1) Linear charge density of the shell: $-2.6\mu C/m$

2) x-component of the electric field at r = 8.7 cm: $1.16\cdot 10^6 N/C$ outward

3) y-component of the electric field at r =8.7 cm: 0

4) x-component of the electric field at r = 1.15 cm: $1.28\cdot 10^7 N/C$ outward

5) y-component of the electric field at r = 1.15 cm: 0

Explanation:

The linear charge density of the cylindrical insulating shell can be found by using

$\lambda_2 = \rho A$

where

$\rho = -567\mu C/m^3$ is charge volumetric density

A is the area of the cylindrical shell, which can be written as

$A=\pi(b^2-a^2)$

where

$b=4.7 cm=0.047 m$ is the outer radius

$a=2.7 cm=0.027 m$ is the inner radius

Therefore, we have :

$\lambda_2=\rho \pi (b^2-a^2)=(-567)\pi(0.047^2-0.027^2)=-2.6\mu C/m$

Here we want to find the x-component of the electric field at a point at a distance of 8.7 cm from the central axis.

The electric field outside the shell is the superposition of the fields produced by the line of charge and the field produced by the shell:

$E=E_1+E_2$

where:

$E_1=\frac{\lambda_1}{2\pi r \epsilon_0}$

where

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 8.7 cm = 0.087 m is the distance from the axis

And this field points radially outward, since the charge is positive .

And

$E_2=\frac{\lambda_2}{2\pi r \epsilon_0}$

where

$\lambda_2=-2.6\mu C/m = -2.6\cdot 10^{-6} C/m$

And this field points radially inward, because the charge is negative.

Therefore, the net field is

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}+\frac{\lambda_2}{2\pi \epsilon_0r}=\frac{1}{2\pi \epsilon_0 r}(\lambda_1 - \lambda_2)=\frac{1}{2\pi (8.85\cdot 10^{-12})(0.087)}(8.2\cdot 10^{-6}-2.6\cdot 10^{-6})=1.16\cdot 10^6 N/C$

in the outward direction.

To find the net electric field along the y-direction, we have to sum the y-component of the electric field of the wire and of the shell.

However, we notice that since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, this means that the net field produced by the wire along the y-direction is zero at any point.

We can apply the same argument to the cylindrical shell (which is also infinite), and therefore we find that also the field generated by the cylindrical shell has no component along the y-direction. Therefore,

$E_y=0$

Here we want to find the x-component of the electric field at a point at

r = 1.15 cm

from the central axis.

We notice that in this case, the cylindrical shell does not contribute to the electric field at r = 1.15 cm, because the inner radius of the shell is at 2.7 cm from the axis.

Therefore, the electric field at r = 1.15 cm is only given by the electric field produced by the infinite wire:

$E=\frac{\lambda_1}{2\pi \epsilon_0 r}$

where:

$\lambda_1=8.2\mu C/m = 8.2\cdot 10^{-6} C/m$ is the linear charge density of the wire

r = 1.15 cm = 0.0115 m is the distance from the axis

This field points radially outward, since the charge is positive . Therefore,

$E=\frac{8.2\cdot 10^{-6}}{2\pi (8.85\cdot 10^{-12})(0.0115)}=1.28\cdot 10^7 N/C$

For this last part we can use the same argument used in part 4): since the wire is infinite, for the element of electric field $dE_y$ produced by a certain amount of charge $dq$ along the wire there exist always another piece of charge $dq$ on the opposite side of the wire that produce an element of electric field $-dE_y$ , equal and opposite to $dE_y$ .

Therefore, the y-component of the electric field is zero.

Learn more about electric field:

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4 0

3 years ago

A mass M suspended by a spring with force constant k has aperiod T when set into oscillations on Earth. Its period on Mars,whose

olchik [2.2K]

Answer:

C)T

Explanation:

The period of a mass-spring system is:

$T=2\pi\sqrt\frac{m}{k}$

As can be seen, the period of this simple harmonic motion, does not depend at all on the gravitational acceleration (g), neither the mass nor the spring constant depends on this value.

6 0

4 years ago