Answer:
3.86×10⁶ Newton/coulombs
Explaination:
Applying,
E = F/q....................... Equation 1
Where E = Electric Field, F = Force, q = charge.
From the question,
Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs
Substitute these values into equation 1
E = 5.4×10⁻¹/ -1.4×10⁻⁷
E = -3.86×10⁶ Newtons/coulombs
Hence the magnitude of the electric field created by the
negative test charge is 3.86×10⁶ Newton/coulombs
Answer:
The Kinetic Energy decreases. The Total Energy stays the same
Explanation:
The TE stays the same, so if PE increases then KE will decrease.
When is at the end of the runway the velocity of the plane is given by the equation

where s=1800 m is the runway length. Thus
At half runway the velocity of the plane is

Therefore at midpoint of runway the percentage of takeoff velocity is
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