Answer:
a)F=3 x 10⁻⁷ N
b)x=2.405 m
Explanation:
Given that
m₁=295 kg
m₂=595 kg
d= 4.1 m
a)
m₃=63 kg
r=d/2 = 2.05 m
The force between the mass m₁ and m₃
by putting the values
F₁₃=2.94 x 10⁻⁷ N
The force between the mass m₂ and m₃
by putting the values
F₂₃=5.94 x 10⁻⁷ N
The net force F
F= F₂₃- F₁₃
F=5.94 x 10⁻⁷ N-2.94 x 10⁻⁷ N
F=3 x 10⁻⁷ N
b)
Lest take at distance x from mass m₂ net force is zero.
Form above two equation
x²=2.01(4.1-x)²
x=1.42 (4.1-x)
x=5.82 - 1.42x
x=2.405 m
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
=
=
Answer:
Explanation:
Given data:
first case
Distance of electric field from center of sphere is r_1 <R
Electric field at r_1< R
second case
Distance of electric field from centre of sphere is r_1 < 2R
Electric field at r_1< 2R
so, we have
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