Two force of 20N and 40N act at a
1 answer:
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Answer:
Δt = 5.85 s
Explanation:
For this exercise let's use Faraday's Law
emf = - d fi / dt
= B. A
\phi = B A cos θ
The bold are vectors. It indicates that the area of the body is A = 0.046 m², the magnetic field B = 1.4 T, also iindicate that the normal to the area is parallel to the field, therefore the angle θ = 0 and cos 0 =1.
suppose a linear change of the magnetic field
emf = - A
Dt = - A
the final field before a fault is zero
let's calculate
Δt = - 0.046 (0- 1.4) / 0.011
Δt = 5.85 s
Answer:
The stress is calculated as
Solution:
As per the question:
Length of the wire, l = 75.2 cm = 0.752 m
Diameter of the circular cross-section, d = 0.560 mm =
Mass of the weight attached, m = 25.2 kg
Elongation in the wire,
Now,
The stress in the wire is given by:
(1)
Now,
Force is due to the weight of the attached weight:
F = mg =
Cross sectional Area, A =
Using these values in eqn (1):