The final speed of the combined masses is 0.186m/s
According to the law of collision, the sum of the momentum of the bodies before the collision is equal to the momentum after the collision.
Mathematically;
m1u1 - m2u2 = (m1+m1)v
- v is the final speed of the combined masses.
Substituting the given parameters;
0.24(0.6) - 0.26(0.2) = (0.24+0.26)v
0.144 - 0.052 = 0.5v
0.092 = 0.5v
v = 0.092/0.5
v = 0.184m/s
Hence the final speed of the combined masses is 0.186m/s
Learn more on collision here: brainly.com/question/7538238
My teacher taught me that besides the type of wave, wavelength doesn't change unless you shorten the wave, and she used string as an example. Making the string shorter causes the wavelength to decrease so it can't be B. Your answer is A
From Newton’s Third Law, the Law of Conservation of Momentum for an elastic collision is derived as m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
<h3>
Newton's third law</h3>
This law states that action and reaction are equal and opposite. That is the force applied to an object is equal to the reaction received by the object.
F₁₂ = -F₂₁
m₁a₁ = -m₂a₂
m₁v₁/t = -m₂v₂/t
m₁v₁ = -m₂v₂
m₁v₁ + m₂v₂ = 0
<h3>For an elastic collision involving two objects</h3>
The sum of the initial momentum must be equal to sum of final momentum.
- let initial velocity = u
- let final velocity = v
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
Thus, from Newton’s Third Law, the Law of Conservation of Momentum for an elastic collision is derived as m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.
Learn more about conservation of linear momentum here: brainly.com/question/7538238
E=hf C=wavelength*F
E=hC/wavelength
E=(6.626*10^-34)*(3.00*10^8)/670*10^-9
E=(6.626*10^-34)*(3.00*10^8)/450*10^-9
We would need to know the time it took to slow to a stop.