I neeed help im so comfused Points Possible: 1, Points Correct: 0 Which group of numbers is listed from greatest to least? -3, -
1 answer:
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The time interval for which the proton remains in the field is -
Δt = .
We have a proton entering a uniform magnetic field which is in a direction perpendicular to the proton's velocity.
We have to determine time interval during which the proton is in the field.
<h3>What is the magnitude of force on the charged particle moving in a uniform magnetic field?</h3>The magnitude of force on the charged particle moving in a uniform magnetic field is given by -
F = qvB sinθ
According to the question, we have -
Entering Velocity (v) = 20 i m/s
Magnetic field intensity (B) = 0.3 T
Leaving velocity (u) = - 20 j m/s
Now -
The entering and leaving velocity vectors have 90 degrees difference between them. Therefore, only a quarter of distance of the complete circular path of radius 'R' is traced by the proton. Therefore -
d = =
Since, the radius of circular path is not given, we will assume it R.
Therefore, time for which proton remained in the field is -
t =
Hence, the time interval for which the proton remains in the field is -
Δt =
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Answer:
460.52 s
Explanation:
Since the instantaneous rate of change of the voltage is proportional to the voltage in the condenser, we have that
dV/dt ∝ V
dV/dt = kV
separating the variables, we have
dV/V = kdt
integrating both sides, we have
∫dV/V = ∫kdt
㏑(V/V₀) = kt
V/V₀ =
Since the instantaneous rate of change of the voltage is -0.01 of the voltage dV/dt = -0.01V
Since dV/dt = kV
-0.01V = kV
k = -0.01
So, V/V₀ =
V = V₀
Given that the voltage decreases by 90 %, we have that the remaining voltage (100 % - 90%)V₀ = 10%V₀ = 0.1V₀
So, V = 0.1V₀
Thus
V = V₀
0.1V₀ = V₀
0.1V₀/V₀ =
0.1 =
to find the time, t it takes the voltage to decrease by 90%, we taking natural logarithm of both sides, we have
㏑(0.01) = -0.01t
So, t = ㏑(0.01)/-0.01
t = -4.6052/-0.01
t = 460.52 s