A dumbbell-shaped object is composed by two equal masses, m, connected by a rod of negligible mass and length r. If I_1 is the m
1 answer:
Answer:
I2>I1
Explanation:
This problem can be solved by using the parallel axis theorem. If the axis of rotation of a rigid body (with moment of inertia I1 at its center of mass) is changed, then, the new moment of inertia is gven by:
where M is the mass of the object and d is the distance of the new axis to the axis of the center of mass.
It is clear that I2 is greater than I1 by the contribution of the term Md^2.
I2>I1
hope this helps!!
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Answer
In this question we have given,
Height of plane, h1=2000m
Height at which charge concentration is 40C, h2=3000m
Height at which charge concentration is -40C, h3=1000m
charge concentaration, q1=40C
charge concentaration, q2=-40C
let the charge concentrations at height h2 and h3 as point charges
Now we will first find the electric feild on plane due to positive charge q1=40
Here k=8.98755*10^9N.m^2/C^2
q1=40C
put values of k, q1 , h1 and h2 in equation 1
\\
V/m[/tex]
similarly electric feild due to negative charge q2=-40
\\
Total electric feild E at the aircraft is given as
...............(2)
Put values of E1 and E2 in equation2
therefore s Total electric feild E at the aircraft is E= 719004V/m