A bullet weighing 0.08 lb is fired with a horizontal velocity of 1800 ft/s into the lower end of a slender 23-lb bar of length L
1 answer:
Answer: hello attached below is the missing image the slender weight is different from what is in the question here so I worked with 23-Ib as requested in the question
answer
≈ 12.17 Rad/sec
Explanation:
weight of bullet ( Wb ) = 0.08 Ib
horizontal velocity = 1800 ft/s
Slender(Wr) = 23-Ib bar with
length ( L ) = 30
h = 12 inches
Vro = 0
<u>Calculate the angular velocity of the bar immediately after the bullet becomes embedded </u>
attached below is a detailed solution
6.708 = ( 0.05011 + 0.5011 ) w'
w' = 6.708 / 0.55121 ≈ 12.17 Rad/sec
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Answer:
attached below
Explanation:
Answer:
maximum speed for safe vehicle operation = 55mph
Explanation:
Given data :
radius ( R ) = 678 ft
old building located ( m )= 30 ft
super elevation = 0.06
<u>Determine the maximum speed for safe vehicle operation </u>
firstly calculate the stopping sight distance
m = R ( 1 - cos ) ---- ( 1 )
R = 678
m ( horizontal sightline ) = 30 ft
back to equation 1
30 = 678 ( 1 - cos (28.655 *s / 678 ) )
( 1 - cos (28.655 *s / 678 ) ) = 30 / 678 = 0.044
cos = 1.044
hence ; 28.65 * s = 678 * 0.2956
s = 6.99 ≈ 7 ft
next we will calculate the design speed ( u ) using the formula below
S = 1.47 ut + ---- ( 2 )
t = reaction time, a = vehicle acceleration, G1 = grade percentage
assuming ; t = 2.5 sec , a = 11.2 ft/sec^2, G1 = 0
back to equation 2
6.99 = 1.47 * u * 2.5 +
3.675 u + 0.0958 u^2 - 6.99 = 0
u ( 3.675 + 0.0958 u ) = 6.99