Answer:
5.16×10¹⁵ s
Explanation:
Considering the rotational motion of the Earth,
K.E. = ω² where, I = MR²
= MR²ω² ---------(1) where ω = angular velocity of Earth
To find ω,
ω = 2π/T where T = time period of Earth
= 2π/(60×60×24) = 7.27×10^-5 rad/s
Substituting the values in (1),
K.E. = 0.4×(6×10^24)×(6400000)²×(7.27×10^-5)²
= 5.16×10²⁹ J
To calculate possible time,
Time = Energy/ Power
= 5.16×10²⁹ / 10¹⁴
= 5.16×10¹⁵ s
When analyzing inelastic collisions, we need to consider the law of conservation of momentum, which states that the total momentum, p, of the closed system is a constant. In the case of inelastic collisions, the momentum of the combined mass after the collision is equal to the sum of the momentum of each of the initial masses.
p1+p2+...=pf
In our case we only have two masses, which makes our problem fairly simple. Lets plug in the formula for momentum; p=mv.
m1v1+m2v2=(m1+m2)vf
To find the velocity of the combined mass we simply rearrange the terms.
vf=m1v1+m2v2m1+m2
Plug in the values given in the problem.
vf=(3.0kg)(1.4m/s)+(2.0kg)(0m/s)03.0kg+2.0kg
vf=.84m/s
The wave nature of light, due to the experiment having bright and dark bands corresponding to places where you have constructive and destructive interference.
Answer:
The second one is the answer