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:
563.86 N
Explanation:
We know the buoyant force F = weight of air displaced by the balloon.
F = ρgV where ρ = density of air = 1.29 kg/m³, g = acceleration due to gravity = 9.8 m/s² and V = volume of balloon = 4πr/3 (since it is a sphere) where r = radius of balloon = 2.20 m
So, F = ρgV = ρg4πr³/3
substituting the values of the variables into the equation, we have
F = 1.29 kg/m³ × 9.8 m/s² × 4π × (2.20 m)³/3
= 1691.58 N/3
= 563.86 N
Answer:
c. Joints allow the roadway to expand and contract as cars put force on the bridge
Explanation:
The reasons why joints are allowed on roadway is to accommodate the contraction and expansion of the road as cars put force on them.
- Most materials used in making roadways are susceptible to expansion and contraction.
- When a measure of force is applied their length either increases or decreases depending on the type of force.
- To accommodate these changes, joints are placed in roadways
It's a measure of the acceleration
