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:
They will sometimes crash into other plates in the process and will rub while they are moving creating earthquakes
Explanation:
Answer:
Explanation:
given,
mass of wheel(M) = 3 Kg
radius(r) = 35 cm
revolution (ω_i)= 800 rev/s
mass (m)= 1.1 Kg
I_{wheel} = Mr²
when mass attached at the edge
I' = Mr² + mr²
using conservation of angular momentum
Answer:
Explanation:
28 / 70 = 0.3857142... = 0.39 hr
280 / 100 = 2.8 hrs.
(100 - 0) / 10 = 10 m/s²
(60 - 20) / 4 = 10 m/s²
Answer:
Explanation:
Using the principle of moment, assuming the rod is uniform rod of mass 1 kg
the center of mass of the rod will be at 1 m
assuming the system is in equilibrium,
clockwise moment = anticlockwise moment
let the distance of the man shoulder be x from the center of gravity and also is the pivot point
total mass of bucket + mass of honey = 2kg + 3 kg = 5 kg for rear bucket and
2kg + 5 kg = 7 kg for front bucket
( 5kg × ( 1+x)) + ( 1 kg × x) = 7 kg × ( 1 - x)
5 + 5 x + x = 7 - 7x
5 + 6x = 7 - 7x
6x + 7x = 7 - 5
13x = 2
x = 2 / 13 = 0.154 m
the honeybucket man's shoulder is 0.154 m from the center of the pole ( forward ).
