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.
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Answer:
d = 0.71 meters
Explanation:
It is given that,
Charge 1, 
Charge 2, 
Electrostatic force between charges, F = 9 N
Let d is the distance between the charges. The electrostatic force between the charges is given by the product of charges and divided by square of distance between them. Mathematically, it is given by :



d = 0.71 meters
So, the distance between the charges is 0.71 meters. Hence, this is the required solution.
Answer:
Velocity has a direction associated with it, while speed has no specific direction.
Explanation:
Velocity is a vector, while speed is a scalar.
From the choices given, ultraviolet rays is a type electromagnetic radiation that causes certain substances to fluoresce. It cannot be detected by the naked eye but some insects are able to see them.
Like in a fluorescent light bulb, ultraviolet or UV lights stimulates the coating of the tube to emit light.
The 3rd option if not sorry I think that is the answer