A string that is under 50.0N of tension has linear density 5.0g/m. A sinusoidal wave with amplitude 3.0cm and wavelength 2.0m travels along the string. What is the maximum speed of a particle on the string

9.42 m/s

Explanation:

Applying,

V' = Aω Equation 1

Where V' = maximum speed of the string, A = Amplitude of the wave, ω = angular velocity.

But,

ω = 2πf Equation 2

Where f = frequency, π = pie

And,

f = v/λ Equation 3

Where, λ = wave length, v = velocity

Also,

v = √(T/μ) Equation 4

Where T = Tension, μ = linear density.

From the question,

Given: T = 50.0 N, μ = 5.0 g/m = 0.005 kg/m

Substitute into equation 4

v = √(50/0.005)

v = √(10000)

v = 100 m/s

Also Given: λ = 2.0 m

Substitute into equation 3

f = 100/2

f = 50 Hz.

Substitute the value of f into equation 2

Where π = constant = 3.14

ω = 2(3.14)(50)

ω = 314 rad/s

Finally,

Given: A = 3.0 cm = 0.03 m

Substitute into equation 1

V' = 0.03(314)

V' = 9.42 m/s

real, smaller and upside-down.   so the answer should be b.

c) up is the direction the magnetic force is acting on the charge.