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Answer:
The acceleration of the proton is 1.403 x 10⁹ m/s²
Explanation:
Given;
speed of proton, v = 7.7 m/s
magnitude of magnetic field, B = 1.9 T
Magnetic force of moving proton is given by;
F = qvBsinθ
Centripetal force on the moving proton is given by;
where;
q is charge of the proton = 1.602 x 10⁻¹⁹ C
m is mass of proton = 1.67 x 10⁻²⁷ kg
Therefore, the acceleration of the proton is 1.403 x 10⁹ m/s²
Refer to the diagram shown below.
The force, F, is applied at 5 cm from the elbow.
For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N
Answer: 1500 N
The force, F, is applied at 5 cm from the elbow.
For dynamic equilibrium, the sum of moments about the elbow is zero.
Take moments about the elbow.
(5 cm)*(F N) - (30 cm)*(250 N) = 0
F = (30*250)/5 = 1500 N
Answer: 1500 N