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Sagot :
[tex]velocity=\frac{distance}{time}\\\\
velocity*time=distance\\\\
time=\frac{distance}{velocity}\\\\
time=\frac{0,840m}{521\frac{m}{s}}=0,0016s\\\\
acceleration=\frac{velocity}{time}=\frac{521}{0,0016}=325625\frac{m}{s^2}[/tex]
Answer:
a = 1.62*105 m /s2
Explanation:
25.28 m2/s2 = vi2
vi = 5.03 m/s
To find hang time, find the time to the peak and then double it.
vf = vi + a*t
0 m/s = 5.03 m/s + (-9.8 m/s2)*tup
-5.03 m/s = (-9.8 m/s2)*tup
(-5.03 m/s)/(-9.8 m/s2) = tup
tup = 0.513 s
hang time = 1.03 s
Return to Problem 11
Given:
vi = 0 m/s
vf = 521 m/s
d = 0.840 m
Find:
a = ??
vf2 = vi2 + 2*a*d
(521 m/s)2 = (0 m/s)2 + 2*(a)*(0.840 m)
271441 m2/s2 = (0 m/s)2 + (1.68 m)*a
(271441 m2/s2)/(1.68 m) = a
a = 1.62*105 m /s2
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