Explanation:
It is given that,
Displacement of the delivery truck,
(due east)
Then the truck moves,
(due south)
Let d is the magnitude of the truck’s displacement from the warehouse. The net displacement is given by :


d = 4.03 km
Let
is the direction of the truck’s displacement from the warehouse from south of east.


So, the magnitude and direction of the truck’s displacement from the warehouse is 4.03 km, 37.4° south of east.
The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

<h3>Further explanation</h3>
Let's recall Elastic Potential Energy formula as follows:

where:
<em>Ep = elastic potential energy ( J )</em>
<em>k = spring constant ( N/m )</em>
<em>x = spring extension ( compression ) ( m )</em>
Let us now tackle the problem!

<u>Given:</u>
mass of object = m = 1.25 kg
initial extension = x = 0.0275 m
final extension = x' = 0.0735 - 0.0275 = 0.0460 m
<u>Asked:</u>
kinetic energy = Ek = ?
<u>Solution:</u>
<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>






<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>







<h3>Learn more</h3>
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<h3>Answer details</h3>
Grade: High School
Subject: Physics
Chapter: Elasticity
Answer:
Option (4)
Explanation:
There are two types of collision.
Perfectly elastic collision: the collision in which the momentum and kinetic energy is conserved. There is no loss of energy in other forms of energy.
Perfectly plastic collision: The collision in which the momentum is conserved and kinetic energy is not conserved. The two bodies stick after the collision.
Here, the bullet hits the block and then embedded in the block, it is the example of plastic collision.
Answer:
some one might report you
Explanation:
When the car moves and makes a sound that is louder that when the car is just sitting there