Complete question:
A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find
(a) the force constant of the spring and (b) the amplitude of the motion.
Answer:
(a) the force constant of the spring = 47 N/m
(b) the amplitude of the motion = 0.292 m
Explanation:
Given;
mass of the spring, m = 200g = 0.2 kg
period of oscillation, T = 0.410 s
total mechanical energy of the spring, E = 2 J
The angular speed is calculated as follows;
(a) the force constant of the spring
(b) the amplitude of the motion
E = ¹/₂kA²
2E = kA²
A² = 2E/k
Since the temperature decreases (cold) , the pressure exerted will decrease. the particles are move slower and collide less vigorously and frequently with the inner surface of the container . therefore since p = f/a , the pressure will decrease as force exerted decreases .
Explanation:
Given:
v₀ = 0 m/s
a = 2.50 m/s²
t = 4 s
Find: v
v = at + v₀
v = (2.50 m/s²) (4 s) + 0 m/s
v = 10 m/s
Answer:
τ=0.060 N.m
Explanation:
By kinematics:
Solving for α:
where ωo = 600*2*π/60; ωf = 0; t=10s
The sum of torque is:
Total work done is 0.13 Joules
<h3>What is work done ?</h3>
The sum of the displacement and the component of the applied force of the object in the displacement direction is the work done by a force.
According to the given information
We need to find the work done
work done = force × distance
We are given,
force = 26 N
Distance = 0.0005 meter
hence ,
Work done = 26 × 0.005
= 0.13 Joules
Total work done is 0.13 Joules
To know more about Work done
brainly.com/question/13662169
#SPJ4
