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
Um this doesn't make since to me since you did not clearly state your awnser
Answer:
B)
Explanation:
That the time period of which they stop.
Answer:
m = 20,000 kg
Explanation:
Force, 
Acceleration of the shark, 
It is required to find the mass of the shark. Let m is the mass. Using second law of motion to find it as follows :
F = ma
Putting the value of F and a to find m

So, the shark's mass is 20,000 kg.
Answer:
Fr = 26.83 [N]
Explanation:
To solve this problem we must use the Pythagorean theorem, since the forces are vector quantities, that is, they have magnitude and density. Therefore the Pythagorean theorem is suitable for the solution of this problem.
![F_{r}=\sqrt{(12)^{2}+(24)^{2} } \\F_{r}=26.83[N]](https://tex.z-dn.net/?f=F_%7Br%7D%3D%5Csqrt%7B%2812%29%5E%7B2%7D%2B%2824%29%5E%7B2%7D%20%20%7D%20%5C%5CF_%7Br%7D%3D26.83%5BN%5D)