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.
Answer:
ΔL = MmRgt / (2m + M)
Explanation:
The system starts from rest, so the change in angular momentum correlates directly to its final angular momentum.
ΔL = L − L₀
ΔL = Iω − 0
ΔL = ½ MR²ω
To determine the angular velocity ω, begin by drawing a free body diagram for both the pulley and the block.
For the block, two forces act: the weight force mg downward and tension force T upward.
For the pulley, three forces are present: weight force Mg down, a reaction force up, and tension force T downward.
For the sum of forces in the -y direction on the block:
∑F = ma
mg − T = ma
T = mg − ma
For the sum of torques on the pulley:
∑τ = Iα
TR = (½ MR²) (a/R)
T = ½ Ma
Substituting gives:
mg − ma = ½ Ma
2mg − 2ma = Ma
2mg = (2m + M) a
a = 2mg / (2m + M)
The angular acceleration of the pulley is:
αR = 2mg / (2m + M)
α = 2mg / (R (2m + M))
Finally, the angular velocity after time t is:
ω = αt + ω₀
ω = 2mg / (R (2m + M)) t + 0
ω = 2mgt / (R (2m + M))
Substituting into the previous equations gives:
ΔL = ½ MR² × 2mgt / (R (2m + M))
ΔL = MmRgt / (2m + M)
The vehicle experiences a normal force of 4440 N. The normal force acts perpendicular to the ground surface. Key details include the vehicle's mass of 1200 kg and the gravitational force of 3.7 N/kg. We calculate the normal force in Newtons by multiplying these two figures: force = field strength * mass = 3.7 N/kg * 1200 kg = 4440 N.
<span>A force of 110 N is applied at an angle of 30</span>°<span> to the horizontal. Because the force does not align directly either vertically or horizontally with the sled, it can be broken down into two components based on sine and cosine.
For the component parallel to the ground:
x = rcos</span>β
<span>x = 110cos30</span>°
<span>x = 95.26
For the component perpendicular to the ground:
y = rsin</span>β
<span>y = 110sin30</span>°
<span>y = 55</span>
The final mass will be slightly lower due to evaporation. I learned this back in third grade, so it's surprising you're in high school and don't know this.
