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
Answer: Different types of telescopes usually don't take simultaneous readings. Space is a dynamic system, so an image taken at one time is not necessarily the precise equivalent of an image of the same phenomena taken at a later time. And often, there is barely enough time for one kind of telescope to observe extremely short-lived phenomena like gamma-ray bursts. By the time other telescopes point to the object, it has grown too faint to be detected.
Explanation: Trust me
We will assume that the CM of the arm is at "L"
from the elbow, and the ball is at 34cm. Then the net torque is computed
by:
Net τ = 1.42 N * 34 cm + 1.50 kg * 9.8m/s² * 34 cm/2 – 12.6 N*2.75cm
= 48.28 N*cm + 1.50kg * 9.8m/s² * 34 cm/2 – 12.6 N*2.75cm
= 48.28 N*cm + 499.8 – 34.65 N*cm
τ = 513.43 N*cm or
5.1343 N*m
Examples of Newton's Second Law of Motion
Pushing a Car and a Truck. ...
Pushing a Shopping Cart. ...
Two People Walking Together. ...
Hitting a Ball. ...
Rocket Launch. ...
Car Crash. ...
Object thrown from a Height. ...
Karate Player Breaking Slab of Bricks.