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Suppose that a particle accelerator is used to move two beams of particles in opposite directions. In a particular region, electrons move to the right at 6020 m/s and protons move to the left at 1681 m/s. The particles are evenly spaced with 0.0476 m between electrons and 0.0662 m between protons. Assuming that there are no collisions and that the interactions between the particles are negligible, what is the magnitude of the average current in this region

Answers

Answer 1

Answer:

Answer:

$2.429783984* 10^(-14)\ A$

Explanation:

Velocity of electron = 6020 m/s

Velocity of proton = 1681 m/s

Electron space = 0.0476 m

Proton space = 0.0662 m

e = Charge of particle = $1.6* 10^(-19)\ C$

Number of electrons passing per second

$n_e=(6020)/(0.0476)\n\Rightarrow n_e=126470.588$

Number of protons passing per second

$n_p=(1681)/(0.0662)\n\Rightarrow n_p=25392.749$

Current due to electrons

$I_e=n_ee\n\Rightarrow I_e=126470.588* 1.6* 10^(-19)\n\Rightarrow I_e=2.0235* 10^(-14)\ A$

Current due to protons

$I_p=n_pe\n\Rightarrow I_p=25392.749* 1.6* 10^(-19)\n\Rightarrow I_p=4.06283984* 10^(-15)\ A$

Total current

$I=2.0235* 10^(-14)+4.06283984* 10^(-15)\n\Rightarrow I=2.429783984* 10^(-14)\ A$

The average current is $2.429783984* 10^(-14)\ A$

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Calculate the work done by a 50.0 N force on an object as it moves 9.00 m, if the force is oriented at an angle of 135° from the direction of travel. Explain what your answer means in terms of the object’s energy.

Answers

Answer:

Work done, W = -318.19 Joules

Explanation:

It is given that,

Force acting on the object, F = 50 N

Distance covered by the force, d = 9 m

Angle between the force and the distance traveled, $\theta=135^(\circ)$

The work done by an object is equal to the product of force and distance traveled. It is equal to the dot product of force and the distance. Mathematically, it is given by :

$W=Fd\ cos\theta$

$W=50* 9* \ cos(135)$

W = -318.19 Joules

So, the work done by the force is 318.19 Joules. The work is done in opposite to the direction of motion. Hence, this is the required solution.

Air contained in a rigid, insulated tank fitted with a paddle wheel, initially at 300 K, 2 bar, and a volume of 2 m3 , is stirred until its temperature is 500 K. Assuming the ideal gas model for the air, and ignoring kinetic and potential energy, determine (a) the final pressure, in bar, (b) the work, in kJ, and (c) the amount of entropy produced, i

Answers

Final answer:

To find the final pressure, use the ideal gas law equation PV = nRT, where P is the initial pressure, V is the initial volume, n is the number of moles of gas, R is the gas constant, and T is the initial temperature. Rearrange the equation and plug in the given values to find that the final pressure is 3.33 bar.

Explanation:

To find the final pressure, we can use the ideal gas law equation: PV = nRT, where P is the initial pressure, V is the initial volume, n is the number of moles of gas, R is the gas constant, and T is the initial temperature.

Since the volume and the amount of air are constant, we can rearrange the equation to solve for the final pressure:

P2 = P1 * (T2 / T1),

where P2 is the final pressure, T2 is the final temperature, and T1 is the initial temperature.

By plugging in the values from the problem, we can find that the final pressure is 3.33 bar.

Learn more about Ideal Gas Law here:

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Jerome is learning how the model of the atom has changed over time as new evidence was gathered. He has images of four models of the atom, but they are not in the correct order.

Answers

Answer:

Y, X, Z, W

Explanation:

Jerome must put the given models in the order Y, X, Z, W to display the development of atom from the earliest to the most recent one. 'Y' represents 'Thomson's plum pudding model' came in 1904 which was followed by the 'Rutherford's nuclear atomic model' of 1911 as represented by X. This was succeeded by the 'Bohr's electrostatic model' in 1913(as shown in model Z) and lastly, the model W which exemplifies the 'Quantum Mechanical Model' by Edwin Schordinger in 1926. Thus, the correct order is Y, X, Z, W.

Answer:YXZW

Explanation:

Temperature°F = (9/5 * °C) + 32°
°C = 5/9 * (°F - 32°)
1 pt each. Using the table above as a guide, complete the following conversions. Be sure to show your work to the side:
1. 5 cm = ________ mm
2. 83 cm = ________ m
3. 459 L = _______ ml
4. .378 Kg = ______ g
5. 45°F = ________ °C
6. 80°C = _________ °F

Answers

$5cm = 50mm$
$2.83cm = 0.0283m$
$3.459l = 3459ml$
$4.378kg = 4378g$
$5.45f = - 47.79c$
$6.80c = 44.24f$

Ted Clubber Lang. A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at the elbow. During this punch, the horizontal flexor muscles of the shoulder contract and shorten at an average speed of 75 cm/s. They move through an arc length of 5 cm during the hook, while the first moves through an arc length of 100 cm. What is the average speed of the first during the hook?

Answers

Answer:

15 m/s or 1500 cm/s

Explanation:

Given that

Speed of the shoulder, v(h) = 75 cm/s = 0.75 m/s

Distance moved during the hook, d(h) = 5 cm = 0.05 m

Distance moved by the fist, d(f) = 100 cm = 1 m

Average speed of the fist during the hook, v(f) = ? cm/s = m/s

This can be solved by a very simple relation.

d(f) / d(h) = v(f) / v(h)

v(f) = [d(f) * v(h)] / d(h)

v(f) = (1 * 0.75) / 0.05

v(f) = 0.75 / 0.05

v(f) = 15 m/s

Therefore, the average speed of the fist during the hook is 15 m/s or 1500 cm/s

Two particles, one with charge -6.29 × 10^-6 C and one with charge 5.23 × 10^-6 C, are 0.0359 meters apart. What is the magnitude of the force that one particle exerts on the other?