Answer:
E = 10t^2e^-10t Joules
Explanation:
Given that the current through a 0.2-H inductor is i(t) = 10te–5t A.
The energy E stored in the inductor can be expressed as
E = 1/2Ll^2
Substitutes the inductor L and the current I into the formula
E = 1/2 × 0.2 × ( 10te^-5t )^2
E = 0.1 × 100t^2e^-10t
E = 10t^2e^-10t Joules
Therefore, the energy stored in the inductor is 10t^2e^-10t Joules
First, foremost, and most critically, you must look at the graph, and critically
examine its behavior from just before until just after the 5-seconds point.
Without that ability ... since the graph is nowhere to be found ... I am hardly
in a position to assist you in the process.
Answer:
Hot water rises and cold water sinks is a model of thermal energy transfer by conduction.
The elastic potential energy stored in the car's spring during the process is 3.75 J
<h3>Determination of the spring constant</h3>
From the question given above, the following data were obtained:
K = F/e
K = 15 / 0.5
K = 30 N/m
<h3>Determination of the potential energy</h3>
- Spring constant (K) = 30 N/m
PE = ½Ke²
PE = ½ × 30 × 0.5²
PE = 15 × 0.25
PE = 3.75 J
Therefore, the elastic potential energy stored in the car's spring during the process is 3.75 J
Learn more about energy stored in spring:
brainly.com/question/4280346
