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:
When a man travels from Hilly region to Terai region, his weight gradually increases because the value of g is more at the Terai region than that in hilly region. 3. An object weights 20 N in air and 16 N in liquid, then answer the following questions.
Explanation:
because the value of g is more at the Terai region than that in hilly region. 3. An object weights 20 N in air and 16 N in liquid, then answer the following questions.
Explanation:
Speed: distance/time
Average speed: total distance/total time
Total distance: 400
Total time: 60
Average spead: 400/60= 6.67m/s
Galaxy million of star and planet. gravitional wave field all the universe some planet explosive itself moving other places . Black holes Mass gravity field
Answer:
Distance, d = 192 meters
Explanation:
We have,
Initial velocity of an object is 10 m/s
Acceleration of the object is 3.5 m/s²
Time, t = 8 s
We need to find the distance travelled by the object during that time. Second equation of motion gives the distance travelled by the object. It is given by :
So, the distance travelled by the object is 192 meters.
