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Answer:
The acceleration of the proton is 9.353 x 10⁸ m/s²
Explanation:
Given;
speed of the proton, u = 6.5 m/s
magnetic field strength, B = 1.5 T
The force of the proton is given by;
F = ma = qvB(sin90°)
ma = qvB
where;
m is mass of the proton, = 1.67 x 10⁻²⁷ kg
charge of the proton, q = 1.602 x 10⁻¹⁹ C
The acceleration of the proton is given by;
Therefore, the acceleration of the proton is 9.353 x 10⁸ m/s²
Answer:
The bit take to reach its maximum speed of 8,42 x10^4 rad/s in an amount of 1.097 seconds.
Explanation:
ω1= 1.72x10^4 rad/sec
ω2= 5.42x10^4 rad/sec
ωmax= 8.42x10^4 rad/sec
θ= 1.72x10^4 rad
α=7.67 x10^4 rad/sec²
t= ωmax / α
t= 8.42 x10^4 rad/sec / 7.67 x10^4 rad/sec²
t=1.097 sec