Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
<em>im confused hold on imma send you a link to the answer</em>Explanation:
Answer:
9213 J
Explanation:
Change in Kinetic energy = Change in Potential energy
= 12,928J - 3715J
=9213 J
For more assistance: +1 (304) 223-3136
Answer:
1.6 m
Explanation:
Given that the launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m.
The time for landing should be calculated by using the second equation of motion formula
h = Ut + 1/2gt^2
Let U = 0
0.5 = 1/2 × 9.8 × t^2
0.5 = 4.9t^2
t^2 = 0.5 / 4.9
t^2 = 0.102
t = 0.32 s
The target should be placed so that the toy car lands on it at:
Distance = 5 × 0.32
distance = 1.597 m
Distance = 1.6 m
Therefore, the target should be placed so that the toy car lands on it 1.6 metres away.
