Answer:
a) a = 4.9 m/s²
b) a = 1.5 m/s²
Explanation:
no friction
F = ma
gsinθ = ma
a = gsinθ
a = 9.8sin30
a = 4.9 m/s²
friction
gsinθ - μmgcosθ = ma
a = g(sinθ - μcosθ)
a = 9.8(sin30 - 0.4cos30)
a = 1.5051...
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
56 Newtons bc w=F×D so if you divide by D on both side you get w/D=F
Spinning top follow the classical mechanics so no space quantization is observed.
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