How should you approach a dock when the wind or current is pushing you away from the dock?
1 answer:
You might be interested in
Answer:
A) 10 m/s
Explanation:
We know that according to conservation of momentum,
m1v1 + m2v2 = m1u1 + m2u2 ..............(equation 1)
where m1 and m2 are masses of two bodies, v1 and v2 are initial velocity before collision and u1 and u2 are final velocities after collision respectively.
From the given data
If truck and car are two bodies
truck : m1 = 2000 Kg v1 = 5 m/s u1 = 0
car : m2 = 1000 kg v2 = 0 u2 = ?
final velocity of truck and initial velocity of car are static because the objects were at rest in the respective time.
substituting the values in equation 1, we get
(2000 x 5) + 0 = 0 + (1000 x u2)
u2 = x 5
= 10 m/s
Hence after collision, car moves at a velocity of 10 m/s
<span>{c o {a la manzana o
<span>Vapor de agua en el aire en una ducha de agua caliente</span>
</span>
<span>Vapor de agua en el aire en una ducha de agua caliente</span>
</span>