Complete Question
The speed of a transverse wave on a string of length L and mass m under T is given by the formula

If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string
Answer:

Explanation:
From the question we are told that
Speed of a transverse wave given by

Maximum Tension is 
Generally making
subject from the equation mathematically we have




Therefore the Linear mass in terms of Velocity is given by

Radiant energy is the energy of electromagnetic and gravitational radiation
Answer
given,
v = 128 ft/s
angle made with horizontal = 30°
now,
horizontal component of velocity
vx = v cos θ = 128 x cos 30° = 110.85 ft/s
vertical component of velocity
vy = v sin θ = 128 x sin 30° = 64 m/s
time taken to strike the ground
using equation of motion
v = u + at
0 =-64 -32 x t
t = 2 s
total time of flight is equal to
T = 2 t = 2 x 2 = 4 s
b) maximum height
using equation of motion
v² = u² + 2 a h
0 = 64² - 2 x 32 x h
64 h = 64²
h = 64 ft
c) range
R = v_x × time of flight
R = 110.85 × 4
R = 443.4 ft