If time is specified, the distance may be estimated in constant acceleration using the formula: X=(at2)/2 if the beginning velocity is 0. (A automobile begins from a stop...) As a result, X=(6*10*10)/2=600/2 = 300 m.
1) v=u+gt = -5 +10*2=15 m/s
2) r=ut+gt^2/2=-5*2+10*2^2/2=-10+20=10m
3) r of heli= vt =5*2=10m
So 10m by bag from answer number 2 plus additional 10m by helicopter equals 20m
Answer:
(a) W= 44N
(b)W= 31.65 N
Explanation:
Data
T=44 N : Maximum force that the rope can withstand without breaking
Newton's second law:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
(a) We apply the formula (1) at constant speed , then, a=0
W: heaviest fish that can be pulled up vertically
∑F = 0
T-W =0
W = T
W= 44N
(b) We apply the formula (1) , a= 1.26 m/s²
W: heaviest fish that can be pulled up vertically
W= m*g
m= W/g
g= 9.8 m/s² : acceleration due to gravity
∑F = 0
T-W = m*a
T= W+(W/g)*a
44=W*(1+1/9.8)* (1.26 )
44= W* 1.39
W= 44/1.39
W= 31.65 N
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