The velocity of B after elastic collision is 3.45m/s
This type of collision is an elastic collision and we can use a formula to solve this problem.
<h3>Elastic Collision</h3>
The data given are;
- m1 = 281kg
- u1 = 2.82m/s
- m2 = 209kg
- u2 = -1.72m/s
- v1 = ?
Let's substitute the values into the equation.
From the calculation above, the final velocity of the car B after elastic collision is 3.45m/s.
Learn more about elastic collision here;
brainly.com/question/7694106
Answer:
markers are 29.76 m far apart in the laboratory
Explanation:
Given the data in the question;
speed of particle = 0.624c
lifetime = 159 ns = 1.59 × 10⁻⁷ s
we know that; c is speed of light which is equal to 3 × 10⁸ m/s
we know that
distance = vt
or s = ut
so we substitute
distance = 0.624c × 1.59 × 10⁻⁷ s
distance = 0.624(3 × 10⁸ m/s) × 1.59 × 10⁻⁷ s
distance = 1.872 × 10⁸ m/s × 1.59 × 10⁻⁷ s
distance = 29.76 m
Therefore, markers are 29.76 m far apart in the laboratory
1. b or c
2. c
3. a? or d
4.
5. a
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Answer:
v = 88.89 [m/s]
Explanation:
To solve this problem we must use the principle of conservation of momentum which tells us that the initial momentum of a body plus the momentum added to that body will be equal to the final momentum of the body.
We must make up the following equation:
where:
F = force applied = 4000 [N]
t = time = 0.001 [s]
m = mass = 0.045 [kg]
v = velocity [m/s]
![4000*0.001=0.045*v\\v=88.89[m/s]](https://tex.z-dn.net/?f=4000%2A0.001%3D0.045%2Av%5C%5Cv%3D88.89%5Bm%2Fs%5D)
