<span>The de-acceleration or negative acceleration of stopping is what damages bones. The ground is rigid and therefore the change in momentum when striking the ground will be large. On the trampoline, the elasticity of the material means that the momentum changes more slowly, resulting in smaller accelerations.</span>
Answer:
A)6.15 cm to the left of the lens
Explanation:
We can solve the problem by using the lens equation:
where
q is the distance of the image from the lens
f is the focal length
p is the distance of the object from the lens
In this problem, we have
(the focal length is negative for a diverging lens)
is the distance of the object from the lens
Solvign the equation for q, we find
And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is
A)6.15 cm to the left of the lens
Answer:
Incident Command Structure, ICS or ICS-like EOC Structure is familiar and aligns with the on-scene incident organization. ICS or ICS-like EOC Structure is familiar and aligns with the on-scene incident organization.
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
=
=
