Answer:
a. A = 0.1656 m
b. % E = 1.219
Explanation:
Given
mB = 4.0 kg , mb = 50.0 g = 0.05 kg , u₁ = 150 m/s , k = 500 N / m
a.
To find the amplitude of the resulting SHM using conserver energy
ΔKe + ΔUg + ΔUs = 0
¹/₂ * m * v² - ¹/₂ * k * A² = 0
A = √ mB * vₓ² / k
vₓ = mb * u₁ / mb + mB
vₓ = 0.05 kg * 150 m / s / [0.050 + 4.0 ] kg = 1.8518
A = √ 4.0 kg * (1.852 m/s)² / (500 N / m)
A = 0.1656 m
b.
The percentage of kinetic energy
%E = Es / Ek
Es = ¹/₂ * k * A² = 500 N / m * 0.1656²m = 13.72 N*0.5
Ek = ¹/₂ * mb * v² = 0.05 kg * 150² m/s = 1125 N
% E = 13.72 / 1125 = 0.01219 *100
% E = 1.219
Answer: 1.88
Explanation
Applying Snell’s Law, sin(1)/sin(2) = n(2)/n(1), where n is the index of refraction and sin 1 and 2 being of incidence and refracted respectively.
1) sin35/sin24 = n(2)/1.33
2) 1.41 = n(2)/1.33
3) n(2) = 1.41 x 1.33
4) n(2) = 1.88
Hope this helps :)
Angular velocity = (75x2pie)/60
=2.5pie ras^-1
linear velocity(or speed) at end of string, v = radius x angular velocity
v= 0.5 x 2.5pie
v=3.93 ms^-1
tension of string (I beleve is centeral force aplied by string), F= (mv^2)/r
F= (0.2 x 3.93^2)/0.5
F=6.18 N
(sorry if wrong)
N stands for Newton
2- upload the graph..
