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3 years ago

When released , what is the kinetic energy of the 1c charge of the preceding problem if it flies past its starting position?

1 answer:

5 0

A) When a charge is moved in an electric field the work done (W) is calculated as charge*(change in potential). We can write W = q*V or V = W/q = 10/1 = 10V . This voltage is a difference in electric potential between 2 points within the field. If the charge is positive, and positive work is done upon it, then the final position is more positive than the original one.

<span>b) If a charge (Q) is released from rest and falls through a potential difference V, then its gain in energy (KE if no other force acts on the charged body) is q*V = 10J. This is the same as the work done in moving the charge to its new position in part (a), and is an example of the conservation of energy.</span>

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Suppose that the electric field in the Earth's atmosphere is E = 1.16 102 N/C, pointing downward. Determine the electric charge

butalik [34]

Answer:

Electric charge in the earth will be $Q=5.231\times 10^5C$

Explanation:

We have given that E = 116 N/C

Radius of the earth R = 6371 km = 6371000 m

We have to find the electric charge in the earth '

We know that electric field due to charge is given by $E=\frac{1}{4\pi \epsilon _0}\frac{Q}{R^20}=\frac{KQ}{R^2}$ . here K is coulomb's constant

So $116=\frac{9\times 10^9\times Q}{(6371000)^2}$

$Q=5.231\times 10^5C$

So electric charge in the earth will be $Q=5.231\times 10^5C$

5 0

3 years ago

A book is sitting on the dashboard of a car that is stopped at a traffic light as the car starts to move forward the book slides

Zina [86]

Answer:

The book remained in its state of rest before the car started to move forward as no direct force acted on it.

Explanation:

According to Newton's first law of motion, a body will continue in its present state of rest, or if it is in motion, will continue to move with uniform speed in a straight line unless aced upon by an external force. This tendency of a body to remain in its state of reset or uniform motion in a straight line is known as inertia and is directly proportional to the mass of the body. The more massive a body, the more inertia it possesses. Thus Newton's first law is also known as the law of inertia.

Considering the case of the book on the dashboard of a stationary car which suddenly starts to move. While the car is stopped at the traffic light, the dashboard where the book sits and the book are both at rest. When the car begins to move forward, the dashboard moves forward with it. However as the book is not a part of the car, no force is directly acting on it, so the book so it stays at rest due to its inertia.

Therefore, as the car is moving forward, the stationary book appears to move backward from the reference point of the car, sliding off the dashboard.

3 0

2 years ago

Guys help me with this question.

sleet_krkn [62]

Answer:

Explanation:

Brownian motion is a random (irregular) motion of particles e.g smoke particle. The set up in the diagram can be used to observe the motion of smoke.

1. The apparatus used are:

A is a source of light

B is a converging lens

C is a glass smoke cell

D is a microscope

2. The uses of the apparatus are:

A - produces the light required to so as to see clearly the movement of the particles.

B - converges the rays of light from the source to the smoke cell.

C - is made of glass and used for encamping the smoke particles so as not to mix with air.

D - is used for the clear view or observation or study of the motion of the smoke particles in the cell.

4 0

3 years ago

The time required to produce one cycle of a wave is known as the waves

eimsori [14]

The time described above is known as the waves Period.
The time which it takes for a particle to complete one full cycle is known as the period. Period is normally measured in seconds. Frequency on the other hand is the number of cycles which are completed in a given period of time e.g a second. periodic time T is given by reciprocal of frequency (1/f).

6 0

3 years ago

X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to

lys-0071 [83]

Answer:

a) $\Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

b) $\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

c) $E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

Explanation

Part a

For this case we can use the Compton shift equation given by:

$\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)$

For this case we know the following values:

$h = 6.63 x10^{-34} Js$

$m_e = 9.109 x10^{-31} kg$

$c = 3x10^8 m/s$

$\theta = 37$

So then if we replace we got:

$\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm$

Part b

For this cas we can calculate the wavelength of the phton with this formula:

$\lambda_0 = \frac{hc}{E_0}$

With $E_0 = 300 k eV= 300000 eV$

And replacing we have:

$\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm$

And then the scattered wavelength is given by:

$\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm$

And the energy of the scattered photon is given by:

$E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV$

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

$E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV$

3 0

3 years ago

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