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rosijanka [135]
3 years ago
13

KATZPSEF1 30.P.041. My Notes Ask Your Teacher A charged particle enters a region of space with a uniform magnetic field B = 1.98

i T. At a particular instant in time, it has velocity v = (1.46 ✕ 106 i + 2.42 ✕ 106 j) m/s. Based on the observed acceleration, you determine that the force acting on the particle at this instant is F = 1.60 k N. What are the following?(a) the sign of the charged particle
(b) the magnitude of the charged particle

Physics
1 answer:
sladkih [1.3K]3 years ago
6 0

Answer: The charge on the particle is positive

While the magnitude = 0.00028C

Explanation:

Please find the attached file for the solution

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To solve this problem, let us recall that the formula for gases assuming ideal behaviour is given as:

rms = sqrt (3 R T / M)

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R = gas constant = 8.314 Pa m^3 / mol K

T = temperature

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Now we get the ratios of rms of Argon (1) to hydrogen (2):

rms1 / rms2 = sqrt (3 R T1 / M1) / sqrt (3 R T2 / M2)

or

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rms1 / rms2 = sqrt (T1 M2 / T2 M1)

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and M2 = 2 while M1 = 40

rms1 / rms2 = sqrt (4 * 2 / 40)

rms1 / rms2 = 0.447

 

Therefore the ratio of rms is:

<span>rms_Argon / rms_Hydrogen = 0.45</span>

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Answer:

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In a television set, electrons are accelerated from rest through a potential difference of 20 kV. The electrons then pass throug
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Answer: Fmax = 5.54*10^-12 N

Explanation: From the question, we have the potential difference (V) =20kv = 20,000v and strength of magnetic field (B) =0.41 T.

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Fmax = 5.54*10^-12 N

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3 years ago
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