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3 years ago

While solving a problem, Tran calculates an answer that has coulomb-volts as the units.

Physics

1 answer:

Bess [88]3 years ago

4 0

Apparently, the question is looking for A. electric potential energy;

but I don't think that's quite right. Electric potential difference is expressed in Joules / Coulomb which is the work to move a charge between 2 points

Example: If the electric field between, say, between 2 capacitor plates is

E = 100 Newtons / Coulomb then the work done in moving a unit of charge from the negative plate to the positive plate separted by 1 cm is

V = E * d = 100 Newtons / Coulomb * .01 meters = 1 Newton-meter / Coulomb

= 1 Joule / Coulomb which is the electric potential or potential difference

(The definition of electric potential between points is "the work moving a unit positive test charge from one point to the other")

Now in our above example where V = 1 Joule / Coulomb

if we move 10 Coulombs from the negative plate to the positive plate

W = V Q = 1 Joule / Coulomb * 10 Coulombs = 10 Joules

where work done has the correct units of Joules.

Your textbook should help clarify this.

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(a) According to Hooke's Law, the force required to hold any spring stretched x meters beyond its natural length is f(x)=kx. Sup

KengaRu [80]

Answer:

a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

Explanation:

a) The work, measured in joules, is a physical variable represented by the following integral:

$W = \int\limits^{x_{f}}_{x_{o}} {F(x)} \, dx$

Where

$x_{o}$ , $x_{f}$ - Initial and final position, respectively, measured in meters.

$F(x)$ - Force as a function of position, measured in newtons.

Given that $F = k\cdot x$ and the fact that $F = 25\,N$ when $x = 0.3\,m - 0.2\,m$ , the spring constant ( $k$ ), measured in newtons per meter, is:

$k = \frac{F}{x}$

$k = \frac{25\,N}{0.3\,m-0.2\,m}$

$k = 250\,\frac{N}{m}$

Now, the work function is obtained:

$W = \left(250\,\frac{N}{m} \right)\int\limits^{0.05\,m}_{0\,m} {x} \, dx$

$W = \frac{1}{2}\cdot \left(250\,\frac{N}{m} \right)\cdot [(0.05\,m)^{2}-(0.00\,m)^{2}]$

$W = 0.313\,J$

The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.

b) Let be $r(\theta) = 2\cdot \sin 5\theta$ . The area of the region enclosed by one loop of the curve is given by the following integral:

$A = \int\limits^{2\pi}_0 {[r(\theta)]^{2}} \, d\theta$

$A = 4\int\limits^{2\pi}_{0} {\sin^{2}5\theta} \, d\theta$

By using trigonometrical identities, the integral is further simplified:

$A = 4\int\limits^{2\pi}_{0} {\frac{1-\cos 10\theta}{2} } \, d\theta$

$A = 2 \int\limits^{2\pi}_{0} {(1-\cos 10\theta)} \, d\theta$

$A = 2\int\limits^{2\pi}_{0}\, d\theta - 2\int\limits^{2\pi}_{0} {\cos10\theta} \, d\theta$

$A = 2\cdot (2\pi - 0) - \frac{1}{5}\cdot (\sin 20\pi-\sin 0)$

$A = 4\pi$

The area of the region enclosed by one loop of the curve $r(\theta) = 2\cdot \sin 5\theta$ is $4\pi$ .

5 0

3 years ago

Explain why the gravitational acceleration of any object near earth is the same matter what was the mass of the object

Gre4nikov [31]

It takes greater force to accelerate an object that has more mass. But the gravitational force between the Earth and an object is greater when the object hass more mass. It works out just right to make any object with any mass accelerate at the same rate.

5 0

3 years ago

I left a location by 6:38am and arrived a new location by 6: 58am. How do I calculate the time spent?

jolli1 [7]

Answer:20 minutes

Explanation:

06:58am - 06:38am =20 minutes

6 0

3 years ago

Read 2 more answers

Which of the following is/are true? Electromagnetic waves are created by accelerating charges. Electromagnetic waves are transve

Vikki [24]

Answer:

All of the above are true.

Explanation:

(a). true

whenever charge particle move back and froth from its mean position then it will produce oscillating electric and magnetic fields, . so an em wave can be obtain by accelerating charge

(b). true

the electric field and the magnetic field have vibrations in the perpendicular direction along the motion of the wave so electromagnetic wave is a transverse wave. therefore, the EM wave is a Transverse wave

(c) true .

The Electromagnetic wave consists of the two mutually perpendicular electric and magnetic fields and also both fields are perpendicular to the direction of propagation of the wave.

(d) true .

An electromagnetic wave carry energy through vacuum with a speed of $3 \times 10^8 m/s$

so , all of the above are true.

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3 years ago

Hurryyyyyy When using the right-hand rule to determine the direction of the magnetic force on a charge, which part of the hand p

masya89 [10]

When using the right-hand rule to determine the direction of the magnetic force on a charge, which part of the hand points in the direction that the charge is moving? The answer is thumb.

One way to remember this is that there is one velocity, represented accordingly by the thumb. There are many field lines, represented accordingly by the fingers. The force is in the direction you would push with your palm. The force on a negative charge is in exactly the opposite direction to that on a positive charge. Because the force is always perpendicular to the velocity vector, a pure magnetic field will not accelerate a charged particle in a single direction, however will produce circular or helical motion (a concept explored in more detail in future sections). It is important to note that magnetic field will not exert a force on a static electric charge. These two observations are in keeping with the rule that magnetic fields do no work.

6 0

3 years ago

Read 2 more answers