Answer: Brittle
Explanation:
took the test and I chose Soft, Soft is the wrong answer don't choose it. The CORRECT ANSWER IS BRITTLE
Answer:

Explanation:
<u>Elastic Potential Energy
</u>
Is the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.

Given a rubber band of a spring constant of k=5700 N/m that is holding potential energy of PE=8600 J, it's required to find the change of length under these conditions.
Solving for Δx:

Substituting:

Calculating:


To solve this problem, we must remember about the law of
conservation of momentum. The initial momentum mist be equal to the final
momentum, that is:
m1 v1 + m2 v2 = (m1 + m2) v’
where v’ is the speed of impact
Since we are not given the masses of each car m1 and m2,
so let us assume that they are equal, such that:
m1 = m2 = m
Which makes the equation:
m v1 + m v2 = (2 m) v’
Cancelling m and substituting the v values:
50 + 48 = 2 v’
2 v’ = 98
v ‘ = 49 km/h
<span>The speed of impact is 49 km/h.</span>
The scale is a rest scale reads the support for is not enough net force.