Answer:
Explanation:
λ=c x²
c = λ / x²
λ is mass / length
so its dimensional formula is ML⁻¹
x is length so its dimensional formula is L
c = λ / x²
= ML⁻¹ / L²
= ML⁻³
B )
We shall find out the mass of the rod with the help of given expression of mass per unit length and equate it with given mass that is M
The mass in the rod is symmetrically distributed on both side of middle point.
we consider a small strip of rod of length dx at x distance away from middle point
its mass dm = λdx = cx² dx
By integrating it from -L to +L we can calculate mass of whole rod , that is
M = ∫cx² dx
= [c x³ / 3] from -L/2 to +L/2
= c/3 [ L³/8 + L³/8]
M = c L³/12
c = 12 M L⁻³
C ) Moment of inertia of rod
∫dmx²
= ∫λdxx²
= ∫cx²dxx²
= ∫cx⁴dx
= c x⁵ / 5 from - L/2 to L/2
= c / 5 ( L⁵/ 32 +L⁵/ 32)
= (2c / 160)L⁵
= (c / 80) L⁵
= (12 M L⁻³/80)L⁵
= 3/20 ML²
=
=
The lunar lander landed on the moon
Answer:
Explanation:
From the exercise we have:
To find the velocity after 2.4s we need to use the following formula:
The negative sign means that the kit is going down.
