Answer:
30 seconds
Explanation:
A = A02^-(t/hl)
--> ln(A/A0) = -(t/hl)ln2
solving for hl,
hl = -t x ln2 /ln(A/A0)
= -(60 min)xln2/ln(50/200)
= 0.5 min or 30 seconds
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
D
Using the kinetic energy 1/2mv^2 formula
5*10^5 is the answer
According to newton's 3rd law of motion,
For every action, there is equal and opposite reaction. So if we move a body against a rough surface, there were be reaction against the force applied.
So using conservation of energy, we know:
Work done to move a body = Work done against Friction
So, Force applied * distance moved = coefficient of Friction * Normal Reaction * distance moved
For a body moving against a normal surface, Normal Reaction (R) = mg
or, mass * acceleration * distance (s) = ∪ * R * distance(s)
or, mass * (v^2/2s) = ∪ * mass * gravity
Now, s = stopping distance = v²/ 2∪g
so, using given value,∪=0.05,
s = v2/2*0.05*g
We know, g = 10, so s = v²/(2*0.05*10) = v²
where v = initial velocity
