Answer:
a) the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b) the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
Explanation:
Given the data in the question;
as the equation of standing wave on a string is fixed at both ends
y = 2AsinKx cosωt
but k = 2π/λ and ω = 2πf
λ = 4 × 0.150 = 0.6 m
and f = v/λ = 260 / 0.6 = 433.33 Hz
ω = 2πf = 2π × 433.33 = 2722.69
given that A = 2.20 mm = 2.2×10⁻³
so
= A × ω
= 2.2×10⁻³ × 2722.69 m/s
= 5.9899 m/s
therefore, the maximum transverse speed of a point on the string at an antinode is 5.9899 m/s
b)
A' = 2AsinKx
= 2.20sin( 2π/0.6 ( 0.075) rad )
= 2.20 sin( 0.7853 rad ) mm
= 2.20 × 0.706825 mm
A' = 1.555 mm = 1.555×10⁻³
so
= A' × ω
= 1.555×10⁻³ × 2722.69
= 4.2338 m/s
Therefore, the maximum transverse speed of a point on the string at x = 0.075 m is 4.2338 m/s
The rate of flow of electric CHARGE past any point is described in the unit of electric CURRENT ... the Ampere.
Answer:
2.5 m/s²
Explanation:
Acceleration: This can be defined as the rate of change of velocity.
The S.I unit of acceleration is m/s²
For circular motion, the expression for acceleration is given as,
a = ω²r ................ Equation 1
Where a = acceleration of the particle, ω = angular speed of the particle, r = radius of the circular path.
Given: ω = 5 rev/s = 31.42 rad/s, r = 0.10 m.
Substitute into equation 1
a = 5²(0.10)
a = 25(0.10)
a = 2.5 m/s²
Hence the acceleration of the particle = 2.5 m/s²
Hence, none of the option is correct
An arachnid has eight legs.