Answer:
The slower the train is moving, the less are the changes of the magnetic flux, thus the eddy currents become weaker.
Explanation:
A magnetic brakes is not a very efficient way of braking when a train is moving slowly because at low speeds, the changes in the magnetic flux are very less and so it causes the eddy current to become weaker.
Let us find the drag force which is proportional to the velocity of two conducting plates.
The EMF that is induced in the eddy currents are : 
The force which is due to the induced magnetic field is, 
Therefore, 
Here, force is directly proportional to the velocity of the two conducting plates.
Therefore, we can say that when the speed of the train is low, the magnetic flux changes are less and thus the eddy currents are weaker.
The answer is B............
Answer: You can Identify any biases about the answer to the scientific question.
Explanation:
Answer:
The time taken by the airplane to take off, t = 11.46 s
Explanation:
Given data,
The initial velocity of the airplane, u = 24 m/s
The acceleration of the plane, a = 8 m/s
The distance covered until take off, d = 800 m
Using the III equation of motion,
v² = u² +2as
= 24² + 2 x 8 x 800
= 13376
v = 115.65 m/s
Using the first equation of motion,
v = u + at
t = (v-u) / a
= (115.65 - 24) / 8
= 11.46 s
Hence, the time taken by the airplane to take off, t = 11.46 s
Answer:
a) 19.4 m/s
b) 19 m/s
Explanation:
a) In the given question,
the potential energy at the initial point = Ui = 0
the potential energy at the final point = Uf = mgh
the kinetic energy at the initial point = Ki = 1/2 mv₀².
the kinetic energy at the final point = Kf = 0
work done by air= Ea= fh = 0.262 N
Now, using the law of conservation of energy
initial energy= final energy
Ki +Ui = Kf + Uf +Ea
1/2 mv₀² + 0 = 0 + mgh + fh
1/2 mv₀² = mgh + fh
h = v₀²/ 2g (1 +f/w)
calculate m
m= w/g = 5.29 /9.8
= 0.54 kg
h = 20 ²/ (2 x9.80) x (1 0.265/5.29)
h = 19.4 m.
b) 1/2 mv² + 2fh = 1/2 mv₀²
Vg = 19 m/s
