Answer:
0.6983 m/s
Explanation:
k = spring constant of the spring = 0.4 N/m
L₀ = Initial length = 11 cm = 0.11 m
L = Final length = 27 cm = 0.27 m
x = stretch in the spring = L - L₀ = 0.27 - 0.11 = 0.16 m
m = mass of the mass attached = 0.021 kg
v = speed of the mass
Using conservation of energy
Kinetic energy of mass = Spring potential energy
(0.5) m v² = (0.5) k x²
m v² = k x²
(0.021) v² = (0.4) (0.16)²
v = 0.6983 m/s
1.085m
Explanation:
Using
a= lambda/sinစ
Sinစ= (587.5*10^-9) x 0.75*10^-3
= 0.000783
Sinစ=0.875*10^-3/d
0.000783= 0.875/d
d= 1.085m
Answer:
(a)
= 0.22 W
(b)
= 0.056 W
Explanation:
given information:
the mass of piano wire, m = 3.00 g = 0.003 kg
tension, F = 25 N
length, l = 80 cm = 0.8 m
frequency, f = 120 Hz
amplitude, A = 1.6 mm = 0.0016 m
(a) the average power carried by the wave, 
=
(√μF)ω²A²
where,
ω = 2πf = 2π120 = 754
μ = 
= 
= 0.00375 kg/m
thus,
=
(√(0.00375)(25))(754)²(0.0016)²
= 0.22 W
(b) What happens to the average power if the wave amplitude is halved.
based on the equation above, we know that the average power is proportional to the square amplitude. therefore


= 
= 0.056 W
Answer: v = 8.4852813742386 m/s
Explanation:
v = √KE ÷ 1/2 M