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
Answer:
a. 250kg I think it's the right answer. hope it helps:)
Let the unknown distance be xmiles
x/39-x/72=11hr
72x-39x/2808=11hr
33x/2808=11
33x= 30888
x=936miles
U can substitue back to check
at speed of 72mph, he would need 936/72=13hrs
at speed of 39mph, he would need 936/39=24hr
the difference is 24-13=11
The cloud of interstellar dust and gas that forms a star is known as a nebula. With an average surface temperature of about 737K, venus is the hottest planet. Hope this helps, and sorry your question didn't get answered in time.
Answer:
0.25 m
Explanation:
Refraction occurs when the velocity or wavelength of a wave changes at the interface between two media.
We know that refractive index=
Wavelength in medium A/wavelength in medium B = velocity in medium A/velocity in medium B
Let the wavelength of medium B be a
0.5/a = 0.3/0.15
0.5 × 0.15 = 0.3 × a
a= 0.5 × 0.15/0.3
a= 0.25 m
