<h2>
Distance traveled in 1 second after drop is 4.9 m</h2><h2>
Distance traveled in 4 seconds after drop is 78.4 m</h2>
Explanation:
We have s = ut + 0.5at²
For a free falling object initial velocity u = 0 m/s and acceleration due to gravity, g = 9.8 m/s²
Substituting
s = 0 x t + 0.5 x 9.8 x t²
s = 4.9t²
We need to find distance traveled in 1 s and 4 s
Distance traveled in 1 second
s = 4.9 x 1² = 4.9 m
Distance traveled in 4 seconds
s = 4.9 x 4² = 78.4 m
Distance traveled in 1 second after drop = 4.9 m
Distance traveled in 4 seconds after drop = 78.4 m
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
The power dissipated across a component can be calculated through the formula P=I^2xR
Substituting the values in we get P=(0.5)^2x10=2.5W
