The elastic potential energy stored in the stretched spring is 1 J.
<h3>What is Hooke's law?</h3>
Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.
Given that;
Force on the spring = 350 Newton
Distance stretched = 7 centimeters or 0.07 m
Hence;
F = ke
k = F/e = 350 Newton/0.07 m = 5000 N/m
Work done in stretching a spring = 1/2ke^2
= 0.5 × 5000 × (2 × 10^-2)^2 =1 J
Learn more about elastic potential energy: brainly.com/question/156316
Answer:
A
Explanation:
its A because your comparing so comparative
<span>0.0001 km / year or 10^-5 km/year just take 50 km and divide it by 5 million</span>
Answer:
The correct option is 'c':electron,proton,helium nucleus
Explanation:
The De-Broglie's wavelength of particle is given by

Thus we can see that wavelength is inversely related to mass of the particle since 'h' (Plank's constant) and velocity is same for all the particles
Thus we conclude that the the lightest particle will have the most wavelength
Electron being the lightest of the 3 particles will have the largest wavelength thus the correct option is 'c'. Since electron has the largest wavelength followed by proton and the least wavelength among the 3 is of helium.
T=s/v=>t=1500/1,5=1000s
1,5km=1500m