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
The X and Y components of the force are 90.63 Newton and 42.26 Newton respectively.
<u>Given the following data:</u>
To determine the X and Y components of the force:
<h3>The horizontal component (X) of a force:</h3>Mathematically, the horizontal component of a force is given by this formula:
Fx = 90.63 Newton.
<h3>The vertical component (Y) of tensional force:</h3>Mathematically, the vertical component of a force is given by this formula:
Fy = 42.26 Newton.
Read more on horizontal component here: brainly.com/question/4080400