Answer:
(a) Final speed of block = 3.2896 m/s
(b) 6.7350 m/s is the speed of the bullet-block center of mass?
Explanation:
Given that:
Mass of bullet (m₁) = 6.20 g
Initial Speed of bullet (u₁) = 929 m/s
Final speed of bullet (v₁) = 478 m/s
Mass of wooden block (m₂) = 850g
Initial speed of block initial (u₂) = 0 m/s
Final speed of block (v₂) = ?
<u>By the law of conservation of momentum as:</u>
<u>m₁×u₁ + m₂×u₂ = m₁×v₁ + m₂×v₂</u>
6.20×929 + 850×0 = 6.20×478 + 850×v₂
Solving for v₂, we get:
<u>v₂ = 3.2896 m/s</u>
Let the V be the speed of the bullet-block center of mass. So,
V = [m₁* u₁]/[m₁ + m₂] (p before collision = p after collision)
= [6.2 *929]/[5.2+850]
<u>V = 6.7350 m/s
</u>
The final answer is 51.6 microC which is answer choice b:
The answer is "Spring Tide" which comes from the English word springen
<span>the vector diagram should start with a set of x-y axes. From the origin, draw a line down 30 units (miles), and from that point draw a line to the right 18 units. To draw the resultant vector, you simply connect the origin to the final point, making a triangle.</span>
i don't understand sorry ;(