Answer:
(a) 42 N
(b)36.7 N
Explanation:
Nomenclature
F= force test line (N)
W : fish weight (N)
Problem development
(a) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled in at constant speed
We apply Newton's first law of equlibrio because the system moves at constant speed:
∑Fy =0
F-W= 0
42N -W =0
W = 42N
(b) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled with an acceleration whose magnitude is 1.41 m/s²
We apply Newton's second law because the system moves at constant acceleration:
m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity
∑Fy =m*a
m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity
F-W= ( W/9.8 )*a
42-W= ( W/9.8 )*1.41
42= W+0.1439W
42=1.1439W
W= 42/1.1439
W= 36.7 N
Answer: 0.0146m
Explanation: The formula that defines the velocity of a simple harmonic motion is given as
v = ω√A² - x²
Where v = linear velocity, A = amplitude = 1.69cm = 0.0169m, x = displacement.
The maximum speed of a simple harmonic motion is derived when x = A, hence v = ωA
One half of maximum speed = speed of motion
3ωA/2 = ω√A² - x²
ω cancels out on both sides of the equation, hence we have that
A/2 = √A² - x²
(0.0169)/2 = √(0.0169² - x²)
0.00845 = √(0.0169² - x²)
By squaring both sides, we have that
0.00845² = 0.0169² - x²
x² = 0.0169² - 0.00845²
x² = 0.0002142
x = √0.0002142
x = 0.0146m
Answer:
0.1143m
Explanation:
W=f×s
8=70s
make s the subject of the formula
s=8/70
=0.1143m
