After a new aircraft has passed ground and flight tests, the aircraft must receive a series of certificates before it can be mas
1 answer:
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Answer:
(a) 0.345 T
(b) 0.389 T
Solution:
As per the question:
Hall emf,
Magnetic Field, B = 0.10 T
Hall emf,
Now,
Drift velocity,
Now, the expression for the electric field is given by:
(1)
And
Thus eqn (1) becomes
where
d = distance
(2)
(a) When
(b) When
Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).
