Response:
The man's speed is 0.144 m/s
Explanation:
This exemplifies conservation of momentum.
The momentum of the ball prior to being caught must equal the momentum of the man-ball system after catching the ball.
Mass of the ball = 0.65 kg
Mass of the man = 54 kg
Speed of the ball = 12.1 m/s
The momentum of the ball before impact can be calculated as mass multiplied by velocity
= 0.65 x 12.1 = 7.865 kg-m/s
After catching the ball, the momentum of the combined system is
(0.65 + 54)Vf = 54.65Vf
Where Vf denotes their final shared velocity.
Setting the initial momentum equal to the final momentum,
7.865 = 54.65Vf
Vf = 7.865/54.65 = 0.144 m/s
Answer:
Explanation:
at 23 degrees Celsius, the diameter measures 4.511 mm
GIVEN DATA:
diameter of hole = 4.500 mm
T_1 = 23.0 degrees Celsius
T_2 = - 78.0 degrees Celsius
the expansion coefficient of aluminum is 2.4*10^{-5} (degrees Celsius)^{-1}
the diameter at 23 degrees Celsius is stated as
= 4.511 mm
the diameter of the rivet after temperature change is given as
= 0.4511 cm
Let's consider a few possibilities.
1. The lowest velocity of the paratrooper would be just before hitting the ground.
2. Given that the jump originated from a relatively short height, the paratrooper utilized a static line, allowing the parachute to deploy almost instantly after leaping.
Hence, we will convert 100 mi/h to ft/s:
100 mi/h * 5280 ft/mi / 3600 s/h = 146.67 ft/sec.
Based on the first assumption, the maximum distance fallen by the paratrooper would equate to 8 seconds at 146.67 ft/s, translating to
8 s * 146.67 ft/s = 1173.36 ft.
This calculated distance is nearly on par with the jump height, validating both assumptions 1 and 2. Thus, this scenario seems plausible.
Moreover, considering the terminal velocity for a parachutist in a freefall position with limbs spread out typically reaches 120 mi/h, which is slightly above the 100 mi/h mentioned in the article. This as well aligns with the notion of the parachute acting like a flag, adding some air resistance.
