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jahnoibenjamin
12.07.2019 •
Physics
Arocket moves upward, starting from rest with an acceleration of +29.4 for 3.98 s. it runs out of fuel at the end of the 3.98 s but does not stop. m/s2 how high does it rise above the ground?
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Ответ:
consider the motion of rocket until it runs out of fuel
v₀ = initial velocity = 0 m/s
v = final velocity when it runs out of fuel = ?
t = time after which fuel is finished = 3.98 sec
a = acceleration = 29.4 m/s²
Y₀ = height gained when the fuel is finished = ?
using the kinematics equation
v = v₀ + a t
v = 0 + (29.4) (3.98)
v = 117.01 m/s
using the equation
v² = v²₀ + 2 a Y₀
(117.01)² = 0² + 2 (29.4) Y₀
Y₀ = 232.85 m
consider the motion of rocket after fuel is finished till it reach the maximum height.
Y₀ = initial position = 232.85 m
Y = final position at maximum height
v₀ = initial velocity just after the fuel is finished = 117.01 m/s
v = final velocity after it reach the maximum height = 0 m/s
a = acceleration due to gravity = - 9.8 m/s²
using the kinematics equation
v² = v²₀ + 2 a (Y - Y₀)
inserting the values
0² = (117.01)² + 2 (- 9.8) (Y - 232.85)
Y = 931.4 m
Ответ:
Entonces la frecuencia es 7 hz
Explanation:
espero te ayude