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
In this item, we are given with the x-component of the velocity. The y-component is equal to 0 m/s. The time it takes for it to reach the volume can be related through the equation,
d = V₀t + 0.5gt²
Substituting the known values,
225 = (0 m/s)(t) + (0.5)(9.8)(t²)
Simplifying,
t = 6.776 s
To determine the distance of the student from the edge of the building, we multiply the x-component by the calculated time.
range = (12.5 m/s)(6.776 s)
range = 84.7 m
<em>Answer: 84.7 m</em>
Answer:
you are making an observation that uses numbers.
Explanation:
D. gravitational force. is your answer hope this helps
