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
(b) the magnitude of the charged particle
1 answer:
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For a stationary siren on a firehouse is blaring at 81Hz. Assume the speed of sound to be 343m/s, the frequency perceived is mathematically given as'
F=81.721Hz
<h3>What is the frequency perceived by a firefighter racing toward the station at 11km/h?</h3>Generally, the equation for the doppler effect is mathematically given as
Therefore
F=81(343+3.05556)/343
F=81.721Hz
In conclusion, the frequency is
F=81.721Hz
Read more about frequency
Answer:
α = 1930.2 rad/s²
Explanation:
The angular acceleration can be found by using the third equation of motion:
where,
α = angular acceleration = ?
θ = angular displacement = (1500 rev)(2π rad/1 rev) = 9424.78 rad
ωf = final angular speed = 0 rad/s
ωi = initial angular speed = (960 rev/s)(2π rad/1 rev) = 6031.87 rad/s
Therefore,
<u>α = - 1930.2 rad/s²</u>
<u>negative sign shows deceleration</u>