A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched
<u>Explanation:</u>
Work, W = 2 J
Initial distance, 
= 30 cm
Final distance, = 42 cm
Force, F = 30 N
Stretched length, x = ?
We know,
W = 1/2 kΔx²
Δx = 42-30 cm = 12 cm = 0.12 m
2 J = 1/2 k X (0.12)²
k = 277.77 N/m
According to Hooke's law,
F = kx
30 N = 277.77 X x
x = 0.108 m
x = 10.8 cm
A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched.
Answer:
The image to the left (with the disks on it)
Explanation:
Interference in any type of wave can be gotten in two forms, constructive interference, and destructive interference.
The constructive interference is between two waves with the same phase, that is, each crest and trough correspond with the crest and trough of the another getting as result a wave with twice the amplitude of the original one.
The destructive interference is between two waves out of phase, in which the crest of one cancels with the trough of another.
If light passes for a slit it will get a diffraction pattern in a screen, at which each bright pattern corresponds to a crest and a dark pattern to a trough, as a consequence of constructive interference and destructive interference in different points of its propagation to the screen.
The circular shape of the disks can be a representation of the wavefront and how the overlaps make constructive and destructive interference in order to get the diffraction pattern.
Answer:
B
Explanation:
I think this one is correct
I believe the correct response would be true, thermal energy or heat that is produced by friction usually cannot be used to do work.
This would be a negative force because they are going in different directions
