The outlaw that was <span>executed by hanging "in the spring of '25" is identified as the HIGHWAYMAN.
This is one of the characters in the song, "American Remains", sang by The Highwaymen. The group consisted of </span><span>Johnny Cash, Waylon Jennings, Willie Nelson and Kris Kristofferson. Other characters in the song were a sailor, a dam builder, and a pilot of a starship.
</span>
This is the first stanza of the song:
"I was a highwayman. Along the coach roads I did ride
<span>With sword and pistol by my side </span>
<span>Many a young maid lost her baubles to my trade </span>
<span>Many a soldier shed his lifeblood on my blade </span>
<span>The b*stards hung me in the spring of twenty-five </span>
<span>But I am still alive."</span>
Answer:
a.Distance = 150 m
b. Displacement = 50 m
Time lapsed = 5 seconds
Explanation:
a. Distance is the change in the position of an object.
The distance covered by the car = 100 + 50
= 150 m
b. Since displacement is a vector quantity,
Displacement of the car = 100 - 50
= 50 m due east
c. Time elapsed is the time taken for the motion of the car starting from when its starts to when it stops.
Thus, the time elapsed = 4 + 1
= 5 seconds
Answer:
The statement is not correct.
Explanation:
To know if the statement is correct, we shall determine the velocity of the car after 3 s. This is illustrated below.
Data obtained from the question include:
Initial velocity (u) = 0 m/s
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) = 3 s
Final velocity (v) =?
v = u + gt
v = 0 + (9.8 × 3)
v = 0 + 29.4
v = 29.4 m/s
Thus, the velocity of the car after 3 s is 29.4 m/s.
Hence, the statement made by the friend is not correct as the car has a falling velocity of 29.4 m/s after 3 s.
Answer:
Net displacement = 0
Distance traveled = 2PQ <_up and down
Explanation:
Answer:
a= 23.65 ft/s²
Explanation:
given
r= 14.34m
ω=3.65rad/s
Ф=Ф₀ + ωt
t = Ф - Ф₀/ω
= (98-0)×
/3.65
98°= 1.71042 rad
1.7104/3.65
t= 0.47 s
r₁(not given)
assuming r₁ =20 in
r₁ = r₀ + ut(uniform motion)
u = r₁ - r₀/t
r₀ = 14.34 in= 1.195 ft
r₁ = 20 in = 1.67 ft
= (1.667 - 1.195)/0.47
0.472/0.47
u= 1.00ft/s
acceleration at collar p
a=rω²
= 1.67 × 3.65²
a = 22.25ft/s²
acceleration of collar p related to the rod = 0
coriolis acceleration = 2ωu
= 2× 3.65×1 = 7.3 ft/s²
acceleration of collar p
= 22.5j + 0 + 7.3i
√(22.5² + 7.3²)
the magnitude of the acceleration of the collar P just as it reaches B in ft/s²
a= 23.65 ft/s²
