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hectorav6619
20.08.2019 •
Physics
Asmall mailbag is released from a helicopter that is descending steadily at 1.39 m/s. (a) after 5.00 s, what is the speed of the mailbag? (b) how far is it below the helicopter? (c) what are your answers to parts (a) and (b) if the helicopter is rising steadily at 1.39 m/s?
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Ответ:
a) 50.4 m/s
b) 122.5 m
c) 47.61 m/s , 122.5 m
Explanation:
a)
v₀ = initial velocity of the small mailbag = - 1.39 m/s
a = acceleration = - 9.8 m/s²
v = final velocity after "5 sec"
t = time interval = 5 sec
using the equation
v = v₀ + a t
v = - 1.39 + (- 9.8) (5)
v = - 50.4 m/s
speed of mailbag after "5 sec" is 50.4 m/s
b)
D = distance traveled by helicopter in "5 sec "
d = distance traveled by small mailbag in "5 sec"
distance traveled by helicopter in "5 sec " is given as
D = v₀ t
D = (- 1.39) (5)
D = - 6.95 m
distance traveled by small mailbag in "5 sec " is given as
d = v₀ t + (0.5) a t²
d = (- 1.39) (5) + (0.5) (- 9.8) (5)²
d = - 129.45 m
y = distance below the helicopter
distance below helicopter is given as
y = |d - D| = |(- 129.45) - (- 6.95)| = 122.5 m
c)
v₀ = initial velocity of the small mailbag = 1.39 m/s
a = acceleration = - 9.8 m/s²
v = final velocity after "5 sec"
t = time interval = 5 sec
using the equation
v = v₀ + a t
v = 1.39 + (- 9.8) (5)
v = - 47.61 m/s
speed of mailbag after "5 sec" is 47.61 m/s
D = distance traveled by helicopter in "5 sec "
d = distance traveled by small mailbag in "5 sec"
distance traveled by helicopter in "5 sec " is given as
D = v₀ t
D = (1.39) (5)
D = 6.95 m
distance traveled by small mailbag in "5 sec " is given as
d = v₀ t + (0.5) a t²
d = (1.39) (5) + (0.5) (- 9.8) (5)²
d = - 115.55 m
y = distance below the helicopter
distance below helicopter is given as
y = |d - D| = |(- 115.55) - (6.95)| = 122.5 m
Ответ: