A)We know the formula of the angular speed is ω = 2π / TWhere T is the time period.When second hand completes one revolution then the time taken is 60s.So T = 60sThen the angular speed of the second hand is ω= 2π / (60s) = 0.1047 rad/sb)When the minute hand completes one revolution the time taken is T = 1 hr = 3600sThen the angular speed of the minute hand is ω =(2π) / (3600s) = 0.001745 rad/sc)When the hour hand completes one revolution then the timeperiod is T = 12hrs = (12)(3600)sThen the angular speed of the hour hand is ω =(2π) / [(12)(3600)s] = 1.45444 x 10^-4 rad/s
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<span>The
best and most correct answer among the choices provided by the question is B.
Reaches a max height of
8.25 feet after 0.63 seconds</span>
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<span> </span>
Answer: 0.091 m
Explanation:
r = 1/B * √(2mV/e), where
r = radius of their circular path
B = magnitude of magnetic field = 1.29 T
m = mass of Uranium -238 ion = 238 * amu = 238 * 1.6*10^-27 kg
V = potential difference = 2.9 kV
e = charge of the Uranium -238 ion = 1.6*10^-19 C
r = 1/1.29 * √[(2 * 238 * 1.6*10^-27 * 2900) / 1.6*10^-19]
r = 1/1.29 * √(2.21*10^-21 / 1.6*10^-19)
r = 1/1.29 * √0.0138
r = 1/1.29 * 0.117
r = 0.091 m
Therefore, the radius of their circular path is 0.091 m
