Aproton travelling at 35 degree to an external magnetic field of magnitude 0.026 t experiences a force of magnitude 5.5x10^-17 n. hint: use formula f = qvbsineθ (a) calculate the speed of the proton. (b) calculate the kinetic energy of the proton in both joules (j) and electron volts (ev).

a) 2.3 x 10⁴ m/s

b) 4.42 x 10⁻¹⁹ J or 2.8 eV

Explanation:

θ = Angle between the direction of motion of proton and magnetic field = 35

B = magnitude of magnetic field = 0.026 T

F = magnitude of magnetic force =5.5 x 10⁻¹⁷ N

q = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C

v = speed of proton

magnetic force on proton is given as

F = q v B Sinθ

5.5 x 10⁻¹⁷ = (1.6 x 10⁻¹⁹) (0.026) Sin35 v

v = 2.3 x 10⁴ m/s

b)

m = mass of the proton = 1.67 x 10⁻²⁷ kg

Kinetic energy of proton is given as

KE = (0.5) m v²

KE = (0.5) (1.67 x 10⁻²⁷) (2.3 x 10⁴)²

KE = 4.42 x 10⁻¹⁹ J

we also know that, 1 eV = 1.6 x 10⁻¹⁹ J

KE =

KE = 2.8 eV

the focus of lesson 1 is newton's first law of motion - sometimes referred to as the law of inertia. newton's first law of motion is often stated as. an object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

explanation:

co2reflects heat radiation back to earth buning coal and oil emits co2 globe warms