Answer:
270 m/s²
Explanation:
Given:
α = 150 rad/s²
ω = 12.0 rad/s
r = 1.30 m
Find:
a
The acceleration will have two components: a radial component and a tangential component.
The tangential component is:
at = αr
at = (150 rad/s²)(1.30 m)
at = 195 m/s²
The radial component is:
ar = v² / r
ar = ω² r
ar = (12.0 rad/s)² (1.30 m)
ar = 187.2 m/s²
So the magnitude of the total acceleration is:
a² = at² + ar²
a² = (195 m/s²)² + (187.2 m/s²)²
a = 270 m/s²
Answer:

Explanation:
m = Mass of object = 
mg = Weight of object = 20 N
g = Acceleration due to gravity = 
v = Final velocity = 15 m/s
u = Initial velocity = 0
d = Distance moved by the object = 150 m
= Angle of slope = 
f = Force of friction
fd = Work done against friction
The force balance of the system is

The work done against friction is
.
Answer:
0.000507 kg/m
Explanation:
L = Length of string
T = Tension
= Mass density of string
E denotes the E string
D denotes the D String
Frequency is given by

So


The mass density of the E string is 0.000507 kg/m