Let Vpa be the velocity of plane with respect to air and Vp and Va be the velocities of plane and air respectively with respect to ground.
<span>Vpa = Vp - Va </span>
<span>Vp = Vpa + Va [all these quantities are vectors] </span>
<span>so by parallelogram law of vector addition </span>
<span>Vp = (Va^2 + Vpa^2 + 2*Va*Vpa*cos120)^(1/2) </span>
<span>Vp^2 = 33^2 + 140^2 - 33*140 </span>
<span>Vp^2 = 16096 </span>
<span>Vp = 126.763m/s </span>
<span>(2) tan b = (33*sin120)/(140+33/2) </span>
<span>b = 10.34 degrees North of East </span>
<span>(3) component of velocity of plane in east = Vp cos(10.34) = 126.763*1 = 126.7m/s </span>
<span>distance travelled by plane in east = 126.7 * 60*60=456120 m</span>
Answer:
When you in a airplane you are in high altitude and thats where air density drops. Thats where ear pops come from.
Explanation:
If an atom<span>, ion, or molecule is at the </span>lowest possible energy<span> level, it and </span>its electrons<span> are said to be in the ground </span>state<span>. If it is at a higher </span>energy<span> level, it is said to be excited, or any </span>electrons<span> that </span>have<span> higher </span>energy<span> than the ground </span>state<span> are excited.</span>
For number one, use the equation of v= v initial+ at. Use v initial as zero for this situation and 9.8 m/s^2 as the acceleration due to gravity
