Converting
releases
absorbs
cooling
transfer
liberating
The car heads east at an average speed of 50 miles per hour from the intersection point towards East. The truck heads east at an average speed of 60 miles per hour from the intersection point towards South.
The distance of car from the intersection point after t hours is 
.
The distance of truck from the intersection point after t hours is 
.
Since these distances are perpendicular to each other, distance apart d (in miles) at the end of t hours is
Thus the distance apart is 
Best Answer:<span> </span><span>The net angle between the direction of flight of the aircraft and the wind in the opposing direction is 20. THe component opposing the aircraft is 2.4*cos 20 KN
= 2400 * 0.9397 = 2255.2622 N. The distance covered is 120 Km
The work done by the aircraft overcoming the wind is
= 2255.2622 * 120000 = 270631464 = 2.71 x 10^8 N
As the question is trickily worded as : the work done on the plane by the air (wind) the answer is -2.71 x 10^8 J . (fourth option)</span>
Answer:
B
Explanation:
the way to get b you have divion each other
Answer:
Option C is the correct answer.
Explanation:
Considering vertical motion of ball:-
Initial velocity, u = 2 m/s
Acceleration , a = 9.81 m/s²
Displacement, s = 40 m
We have equation of motion s= ut + 0.5 at²
Substituting
s= ut + 0.5 at²
40 = 2 x t + 0.5 x 9.81 x t²
4.9t² + 2t - 40 = 0
t = 2.66 s or t = -3.06 s
So, time is 2.66 s.
Option C is the correct answer.
