This question is incomplete, the complete question is;
Car B is rounding the curve with a constant speed of 54 km/h, and car A is approaching car B in the intersection with a constant speed of 72 km/h. The x-y axes are attached to car B. The distance separating the two cars at the instant depicted is 40 m. Determine: the angular velocity of Bxy rotating frame (ω).
Answer:
the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s
Explanation:
Given the data in the question and image below and as illustrated in the second image;
distance S = 40 m
V
= 54 km/hr
V
= 72 km/hr
α = 100 m
now, angular velocity of Bxy will be;
ω = V / α
so, we substitute
ω = ( 54 × 1000/3600) / 100
ω = 15 / 100
ω = 0.15 rad/s
Therefore, the angular velocity of Bxy rotating frame (ω) is 0.15 rad/s
Answer:
C) 24.4°
Explanation:
let nd = 2.419 be the index of refraction of diamond and na = 1.0 be the index of refraction of air and ∅c be the critical angle.
according to Snell's Law:
sin(∅c) = na/nd
sin(∅c) = (1.0)/(2.419)
∅c = 24.4°
Answer:
A) When the angle between the Force (F) and Displacement (x) is 0°, because, Work done (W) is directly proportional to the Cosine of the Angle between the Force applied and the resultant displacement of the subject.
W = F•x cos ∅
If ∅ = 0°,
W = F•x ===> Maximum Work Done.
If ∅ = 45°,
W = F•x/√2
If ∅ = 90°,
W = 0
If ∅ = 180°,
W = –F•x ===> Minimum Work Done.
Be heavier
density=mass÷volume
if two items have the same size they have the same volume so the heavier one will be the denser one
