Answer:
B. Truck X was ahead, not truck Y.
Explanation:
Let's analyze the information provided.
Truck X moved from the point (0,20) to (2.8,50). This indicates that it began at the 20th kilometer and reached 50 km in 2.8 hours. Thus, its speed is v1 = (s2 - s1) / t
v1 = (50 - 20) / 2.8
v1 = 10.7 km/h
Given that it started from the 20th km, it indeed had a head start. Since the line on the graph is linear, this shows its speed was constant without any change in direction.
On the other hand, Truck Y's movement went from the origin (0,0) to (5,20), meaning it took 5 hours to travel 20 km, resulting in a speed of v2 = 20 / 5
v2 = 4 km/h
Again, the straightness of its graph line signifies it maintained a constant speed in a single direction.
Thus, it is evident that Rosa erred in her assumption that Truck Y had a head start.
The kangaroo reaches a maximum vertical altitude of 2.8 m, which can be calculated using the formula 2.8 = 1/2 * 9.8 * t^2. Thus, applying the equation s = ut + 1/2at^2.
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.
Objects in vertical motion are an illustration of non-uniform motion. At the peak of the circle, centripetal force is balanced by the object's weight. Therefore, the minimum speed required at this top point is given by v =
=
= 5.23 m/s. As the sphere descends from the top to the bottom of the circle, according to the law of conservation of energy, potential energy can be expressed as
, where h signifies the diameter of the circle (2r). Hence, the expression will be written as
where u is the velocity at the lowest point. Consequently, the modified equation is
= 
= 
= 11.71 m/s. The collision of the dart with the bullet is an inelastic one. According to the conservation of momentum: v = 
= 
= 
= 58.55 m/s. Thus, the dart's minimum initial speed for the combined system to complete a circular loop post-collision is 58.55 m/s.
The answer is 9938.8 km. Explanation: 1 pound-force = 4.48 N. Hence, 30.0 pounds-force = 134.4 N. The gravitational force between Earth and an object on its surface is defined by: Where M denotes Earth’s mass, m is the object's mass, and R represents the Earth's radius (6371 km). To determine height (h) above Earth's surface, we compare ratios. Ultimately, Pete's weight would be 30 pounds at a height of 9938.8 km from the Earth's surface.
