Askater holds her arms outstretched as she spins at 150 rpm. what is the speed of her hands if they are 155 cm apart?

Assuming her hands stay in symmetry (Ignore my bad spelling) They have a radius of half off 155 cm, using that information you can use (C = (2x3.14) x Radius) Which gives us 486.95cm, knowing that she rotates 150 times a minute she rotates 2.5 times in a second, we then times 486.95cm by 2.5 getting 1217.375,

So her hands are travelling at 1217.375 cm per second.

Or 1.22 meters a second, if you need it in different states feel free to ask

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