Answer:
= +3,394 103 m / s
Explanation:
We will solve this problem with the concept of the moment. Let's start by defining the system that is formed by the complete rocket before and after the explosions, bone with the two stages, for this system the moment is conserved.
The data they give is the mass of the first stage m1 = 2100 kg, the mass of the second stage m2 = 1160 kg and its final velocity v2f = +5940 m / s and the speed of the rocket before the explosion vo = +4300 m / s
The moment before the explosion
p₀ = (m₁ + m₂) v₀
After the explosion
pf = m₁
+ m₂ 
p₀ = [texpv_{f}[/tex]
(m₁ + m₂) v₀ = m₁
+ m₂
Let's calculate the final speed (v1f) of the first stage
= ((m₁ + m₂) v₀ - m₂
) / m₁
= ((2100 +1160) 4300 - 1160 5940) / 2100
= (14,018 10 6 - 6,890 106) / 2100
= 7,128 106/2100
= +3,394 103 m / s
come the same direction of the final stage, but more slowly
My personal understanding and opinion is that ALL of those questions
should be part of an assessment of Physical Activity Readiness.
The atomic number is the same as the proton number so the answer would be D) 10
Answer:
Explanation:
This is case of interference in thin films
for constructive interference in thin film the condition is
2μ t = (2n+1)λ/2 ; μ is refractive index of oil , t is thickness of oil , λ is wave length of light .
2 x 1.28 x t = λ/2 , if n = 0
2 x 1.28 x t = 605 /2
t = 118.16 nm .
the minimum non-zero thickness of the oil film required = 118.16 nm.