Answer:
<em>1.228 x </em><em> mm </em>
<em></em>
Explanation:
diameter of aluminium bar D = 40 mm
diameter of hole d = 30 mm
compressive Load F = 180 kN = 180 x N
modulus of elasticity E = 85 GN/m^2 = 85 x Pa
length of bar L = 600 mm
length of hole = 100 mm
true length of bar = 600 - 100 = 500 mm
area of the bar A = =
= 1256.8 mm^2
area of hole a = =
= 549.85 mm^2
Total contraction of the bar =
total contraction =
==> = <em>1.228 x </em>
<em> mm </em>
Answer:
The inducerd emf is 1.08 V
Solution:
As per the question:
Altitude of the satellite, H = 400 km
Length of the antenna, l = 1.76 m
Magnetic field, B =
Now,
When a conducting rod moves in a uniform magnetic field linearly with velocity, v, then the potential difference due to its motion is given by:
Here, velocity v is perpendicular to the rod
Thus
e = lvB (1)
For the orbital velocity of the satellite at an altitude, H:
where
G = Gravitational constant
= mass of earth
= radius of earth
Using this value value in eqn (1):