<span>it fairly is going to attain a speed of 24 m/s in a 2d, yet between t = 0 and t = a million, it fairly is not any longer vacationing at that speed, yet at slower speeds. it fairly is 12 meters. ?D = [ ( a?T^2 + 2?Tv_i ) ] / 2 the place: ?D = displacement a = acceleration ?T = elapsed time v_i = preliminary speed ?D = [ ( 24m/s^2 • 1s • 1s + 2 • 1s • 0m/s ) ] / 2 ?D = 24 / 2 ?D = 12m</span>
<h2>
Option 3, 216 m is the correct answer.</h2>
Explanation:
We have initial velocity, u = 15 m/s
Time, t = 12 seconds
Final velocity, v = 21 m/s
We have equation of motion v = u + at
Substituting
21 = 15 + a x 12
a = 0.5 m/s²
Now we have equation of motion v² = u² + 2as
21² = 15² + 2 x 0.5 x s
s = 216 m
Displacement = 216 m
Option 3, 216 m is the correct answer.
Answer:
nucleus is the center of atom
Time taken to reach water :
Now, initial vertical speed , u = 0 m/s.
By equation of motion :
Here, a = g = acceleration due to gravity = 9.8 m/s².
So,
Therefore, the height of the bridge is 3.46 m.
Hence, this is the required solution.
<span> the tangent function of 2mi/5mi = 21.8 degrees
cosine of that angle = cos (21.8 degrees)
=0.928
multiply 0.928 with 570 = 570*0.928
Closing velocity = 529.23 mph
</span><span>As the plane gets closer to radar station, tangent function becomes infinity and the cosine function becomes zero.....
</span>0*570 = 0
