Explanation:
Given that,
Mass of the object, m = 7.11 kg
Spring constant of the spring, k = 61.6 N/m
Speed of the observer, 
We need to find the time period of oscillation observed by the observed. The time period of oscillation is given by :
Time period of oscillation measured by the observer is :
So, the time period of oscillation measured by the observer is 5.79 seconds.
Answer:
The work done against gravity is 78.4 J
Explanation:
The work is calculated by multiplying the force by the distance that the
object moves
W = F × d, where W is the work , F is the force and d is the distance
The SI unit of work is the joule (J)
We need to find the work done against gravity when lowering a
16 kg box 0.50 m
→ F = mg
→ m = 16 kg, and g = 9.8 m/s²
Substitute these value in the rule
→ F = 16 × 9.8 = 156.8 N
→ W = F × d
→ F = 156.8 N and d = 0.50
Substitute these values in the rule
→ W = 78.4 J
<em>The work done against gravity is 78.4 J</em>
Since both heat and work can be measured and quantified, this is the same as saying that any change in the energy of a system must result in a corresponding change in the energy of the surroundings outside the system. In other words, energy cannot be created or destroyed.
I only know what number 1. is and its Mechanical Energy.
