Answer:
The horizontal velocity is constant at 16 m/s.
After 1 sec since v = a t then 9.8 m/s^2 * 1 sec = 9.8 m/s for the vertical velocity
V = (Vx^2 + Vy^2)^1/2 = (16^2 + 9.8^2)^1/2 = 18.8 m/s
Answer:
0.27 kg-m/s
Explanation:
i believe this is the correct answer
The lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².
To find the answer, we need to know about the Newton's equation of motion.
<h3>What's the Newton's equation of motion to find the acceleration in term of initial velocity, final velocity and distance?</h3>
- The Newton's equation of motion that connects velocity, distance and acceleration is V² - U²= 2aS
- V= final velocity, U= initial velocity, S= distance and a= acceleration
<h3>What's the acceleration, if the initial velocity, final velocity and distance are 0 m/s, 360km/h and 1.8 km respectively?</h3>
- Here, S= 1.8 km or 1800 m, V= 360km/h or 100m/s , U= 0 m/s
- So, 100²-0= 2×a×1800
=> 10000= 3600a
=> a= 10000/3600 = 2.8 m/s²
Thus, we can conclude that the lowest constant acceleration needed for takeoff from a 1.80 km runway is 2.8 m/s².
Learn more about the Newton's equation of motion here:
brainly.com/question/8898885
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The Coriolis Effect (due to the spinning of the Earth) <span>causes
moving objects to deviate to the right in the Northern Hemisphere
and to deviate to the left in the Southern Hemisphere.
</span>