Answer:
The minimun height is 242 [m]
Explanation:
We can solve this problem by using the principle of energy conservation, where potential energy becomes kinetic energy. We will take the point where the Falcon reaches the speed of 69 (m/s), as the point where the potential energy is zero, i.e. it will be the reference point.
At the reference point all potential energy has been transformed into kinetic energy, therefore the kinetic energy can be calculated.
![E_{k}=0.5*m*v^{2} \\ where:\\v = velocity = 69 [m/s]\\m = mass = 480[g] = 0.480[kg]\\E_{k} = kinetic energy [J]\\E_{k} =0.5*0.48*(69)^{2} \\E_{k} =1142.64[J]](https://tex.z-dn.net/?f=E_%7Bk%7D%3D0.5%2Am%2Av%5E%7B2%7D%20%5C%5C%20where%3A%5C%5Cv%20%3D%20velocity%20%3D%2069%20%5Bm%2Fs%5D%5C%5Cm%20%3D%20mass%20%3D%20480%5Bg%5D%20%3D%200.480%5Bkg%5D%5C%5CE_%7Bk%7D%20%3D%20kinetic%20energy%20%5BJ%5D%5C%5CE_%7Bk%7D%20%3D0.5%2A0.48%2A%2869%29%5E%7B2%7D%20%5C%5CE_%7Bk%7D%20%3D1142.64%5BJ%5D)
Now we can calculate the elevation with respect to the reference point using the definition of the potential energy.
![E_{p}=m*g*h\\ E_{p}=E_{k} \\therefore\\h= E_{p}/(m*g)\\h= 1142.64/(.48*9.81)\\h=242[m]](https://tex.z-dn.net/?f=E_%7Bp%7D%3Dm%2Ag%2Ah%5C%5C%20E_%7Bp%7D%3DE_%7Bk%7D%20%5C%5Ctherefore%5C%5Ch%3D%20E_%7Bp%7D%2F%28m%2Ag%29%5C%5Ch%3D%201142.64%2F%28.48%2A9.81%29%5C%5Ch%3D242%5Bm%5D)
Let us consider the air with the index 1 and the lucite with index 2. Using the Snell's Secound Law, we have:
Entering the unknowns, remembering that the air refrective index is 1 and the lucite refrective index is 1.5, comes:
Using the arcsin properties, we get:
Obs: Approximate results, and the drawing is attached
If you notice any mistake in my english, let me know, because i am not native.
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>
Answer:
n = 5 approx
Explanation:
If v be the velocity before the contact with the ground and v₁ be the velocity of bouncing back
= e ( coefficient of restitution ) = 
and
h₁ is height up-to which the ball bounces back after first bounce.
From the two equations we can write that
So on
= .00396
Taking log on both sides
- n / 2 = log .00396
n / 2 = 2.4
n = 5 approx
Answer:
the branch of mechanics concerned with the interaction of electric currents with magnetic fields or with other electric currents.
Explanation:
