Aluminum foil reflects more light
The orbital period increases if the orbital distance is increased.
Momentum = mass x velocity
Before collision
Momentum 1 = 2 kg x 20 m /s = 40 kg x m/s
Momentum 2 = 3 kg x -10m/s = -30 kg x m/s
After collision
Momentum 1 = 2 kg x -5 m/s = -10 m/s
Momentum 2 = 3 kg x V2 = 3V2
Total momentum before = total momentum after
40 + -30 = -10 + 3V2
V2 = <span>6.67 m/s
Total kinetic energy before
</span><span>= (1/2) [ 2 kg * 20 m/s * 2 + 3 kg * ( -10 m/s) *2 ]
= 550 J
</span>
<span>Total kinetic energy after
</span>= (1/2) [ 2 kg * ( - 5 m/s) * 2 + 3 kg * 6.67 m/s *2 ]
= 91.73 J
Total kinetic energy lost during collision
=<span>550 J - 91.73 J
= 458.27 J</span>
Angular momentum is conserved, just before the clay hits and just after;
<span>mv(L/2) = Iw </span>
<span>I is the combined moment of inertia of the rod, (1/12)ML^2 , and the clay at the tip, m(L/2)^2 ; </span>
<span>I = [(1/12)ML^2 + m(L/2)^2] </span>
<span>Immediately after the collision the kinetic energy of rod + clay swings the rod up so the clay rises to a height "h" above its lowest point, giving it potential energy, mgh. From energy conservation in this phase of the problem; </span>
<span>(1/2)Iw^2 = mgh </span>
<span>Use the "w" found in the conservation of momentum above; and solve for "h" </span>
<span>h = mv^2L^2/8gI </span>
<span>Next, get the angle by noting it is related to "h" as; </span>
<span>h = (L/2) - (L/2)Cos() </span>
<span>So finally </span>
<span>Cos() = 1- 2h/L = 1 - mv^2L/4gI </span>
<span>m=mass of clay </span>
<span>M=mass of rod </span>
<span>L=length of rod </span>
<span>v=velocity of clay</span>
Answer:
Explanation:
The volume flow rate of a fluid in a pipe is given by:
where
A is the cross-sectional area of the pipe
v is the speed of the fluid
In this problem, at the initial point we have
v = 0.84 m/s is the speed of the water
r = 0.21 m is the radius of the pipe, so the cross-sectional area is
So, the volume flow rate is
