Answer:
0.0605 Kg m^2
Explanation:
In this case where we have find he moment of inertia of this object about an axis perpendicular to the x-y plane and passing through the origin, we can just add three moment of inertia's .
MOI= 0.25×0.3^2 + 0.35×0.4^2- 0.45×0.2^2
= 0.0605 Kg m^2
Scientists use models to study atoms, Because atoms are very tiny, and almost impossible to cut open and look inside.. And the center of an atom is the nucleus and that is ever smaller.
U need to set up n solve the general eqn for simple harmonic motion:
x" = -(k/m)x
solution is x(t) = (x0)*cos(wt) + (v0/w)*sin(wt)
where w=sqrt(k/m), x0 is x-position at t=0 and v0 is vel at t=0
u already calculated f in Q.2 and w = 2*pi*f
x0 is 0 as it starts at eqm
v0 is given at 5.1
so u have x(t)
vel is given by x'(t) = (x0)*(-w)*sin(wt) + (v0/w)*w*cos(wt)
substitute t=0.32, x0=0, v0=5.1 n w in the above, u can solve for v at t=0.32.
Answer:
d = 120 [m]
Explanation:
In order to solve this problem, we must use the theorem of work and energy conservation. Where the energy in the final state (when the skater stops) is equal to the sum of the mechanical energy in the initial state plus the work done on the skater in the initial state.
The mechanical energy is equal to the sum of the potential energy plus the kinetic energy. As the track is horizontal there is no unevenness, in this way, there is no potential energy.
E₁ + W₁₋₂ = E₂
where:
E₁ = mechanical energy in the initial state [J] (units of Joules)
W₁₋₂ = work done between the states 1 and 2 [J]
E₂ = mechanical energy in the final state = 0
E₁ = Ek = kinetic energy [J]
E₁ = 0.5*m*v²
where:
m = mass = 60 [kg]
v = initial velocity = 12 [m/s]
Now, the work done is given by the product of the friction force by the distance. In this case, the work is negative because the friction force is acting in opposite direction to the movement of the skater.
W₁₋₂ = -f*d
where:
f = friction force = 36 [N]
d = distance [m]
Now we have:
0.5*m*v² - (f*d) = 0
0.5*60*(12)² - (36*d) = 0
4320 = 36*d
d = 120 [m]
Do you mean mitosis? If so metaphase.
