An archer puts a 0.4 kg arrow to the bowstring. an average force of 190.4 n is exerted to draw the string back 1.47 m. the acceleration of gravity is 9.8 m/s². assuming no frictional loss, with what speed does the arrow leave the bow? answer in units of m/s. if the arrow is shot straight up, how high does it rise? answer in units of m.
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Ответ:
v = 37.4 m/s , h = 71.39m
Explanation:
To find the velocity given:
m = 0.4 kg
F =190.4 N
d = 1.47 m
g = 9.8 m/s^2
So use the equation of work to solve the kinetic energy
W = F *d = 190.4 N * 1.47m
W = 279.88 J
Ke = 1 / 2 * m* v^2
v = √2*Ke / m =√ 2 *279.88 / 0.4 kg
v = 37.4 m/s
Now to find the high to rise can use the conserved law so:
Ke = Pe
279.88 = m*g*h
Solve to h'
h = 279.88 / 0.4kg * 9.8m/s^2
h =71.39 m
Ответ:
your answer would either be (C), (A), or (D)
Explanation:
im not quite sure