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alexandr402 [8]

3 years ago

7

A 580-mm long tungsten wire, with a 0.046-mm-diameter circular cross section, is wrapped around in the shape of a coil and used

as a filament in an incandescent light bulb. When the light bulb is connected to a battery, a current of 0.526 A is measured through the filament. (Note: tungsten has a resistivity of 4.9 × 10-8 ? • m.) How many electrons pass through this filament in 5 seconds?
How many electrons pass through this filament in 5 seconds? 4.22 × 10-19 2.63 3.29 × 1018 1.64 × 1019

What is the resistance of this filament? What is the resistance of this filament? 4.28 ? 58.5 m? 17.1 ? 9.0 ?

What is the voltage of the battery that would produce this current in the filament? 31 mV 9.0 V 2.6 × 10-8 V 32 V

1 answer:

AleksAgata [21]3 years ago

3 0

Explanation:

Below is an attachment containing the solution.

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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

Read 2 more answers

A 10 kg turkey, He kicks the 0.5 kg ball with a force of 50N for 0.2 seconds and the ball flies straight away horizontally from

Harman [31]

Answer:

a. 20m/s

b.50N

c. Turkey has a larger mass than the ball. Neglible final acceleration and therefore remains stationery.

Explanation:

a. Given the force as 50N, times as 0.2seconds and the weight of the ball as 0.5 kg, it's final velocity can be calculated as:

$F\bigtriangleup t=m\bigtriangleup v\\\\50N\times 0.2s=0.5kg\times \bigtriangleup v\\\\\bigtriangleup v=2(50N\times0.2)\\\\=20m/s$

Hence, the velocity of the ball after the kick is 20m/s

b.The force felt by the turkey:

#Applying Newton's 3rd Law of motion, opposite and equal reaction:

-The turkey felt a force of 50N but in the opposite direction to the same force felt by the ball.

c. Using the law of momentum conservation:

-Due to ther external forces exerted on the turkey, it remains stationery.

-The turkey has a larger mass than the ball. It will therefore have a negligible acceleration if any and thus remains stationery.

-Momentum is not conserved due to these external forces.

5 0

3 years ago

There is a force which resists changes in motion, called

adoni [48]

Answer:

inertia 11

Explanation:

5 0

3 years ago

In some circumstances, it is useful to look at the linear velocity of a point on the blade. The linear velocity of a point in un

mihalych1998 [28]

Answer:

$v=wr$

Explanation:

<u>Tangent and Angular Velocities</u>

In the uniform circular motion, an object describes the same angles in the same times. If $\theta$ is the angle formed by the trajectory of the object in a time t, then its angular velocity is

$\displaystyle w=\frac{\theta}{t}$

if $\theta$ is expressed in radians and t in seconds the units of w is rad/s. If the circular motion is uniform, the object forms an angle $2\theta$ in 2t, or $3\theta$ in 3t, etc. Thus the angular velocity is constant.

The magnitude of the tangent or linear velocity is computed as the ratio between the arc length and the time taken to travel that distance:

$\displaystyle v=\frac{\theta r}{t}$

Replacing the formula for w, we have

$\boxed{ v=wr}$

4 0

3 years ago

1. Compare the speed that light waves travel in air to the speed that sound waves travel in the air. How many times faster do li

Vladimir79 [104]

Answer:

895522 times faster.

Explanation:

From the question given above, the following data were obtained:

Speed of sound in air (v) = 335 m/s

Speed of light in air (c) = 3×10⁸ m/s

How many times faster =.?

To obtain how many times faster light travels in air than sound, do the following

c : v => 3×10⁸ : 335

c/v = 3×10⁸ / 335

c/v = 895522

Cross multiply

c = 895522 × v

From the illustrations made above, we can see that the speed of the light in air (c) is 895522 times the speed of sound in air.

Thus, light travels 895522 times faster than sound in air.

6 0

2 years ago