To solve this problem we will apply the principles of conservation of energy, for which we have to preserve the initial kinetic energy as elastic potential energy at the end of the movement. If said equality is maintained then we can affirm that,


Here,
m = mass
k = Spring constant
x = Displacement
v = Velocity
Rearranging to find the velocity,



Our values are,



Replacing our values we have,


Therefore the velocity is 
Answer:
Option A is correct - While a guitar string is vibrating, you gently touch the midpoint of the string to ensure that the string does not vibrate at that point. The lowest-frequency standing wave that could be present on the string vibrates at twice the fundamental frequency.
Explanation:
Before touching the midpoint of the string, the string vibrates with one loop.
Fundamental frequency, f1 = v/(2*L)
Now, when the midpoint of the guitar string was touched, the string vibrates with two loops.
Hence, f2 = 2*v/(2*L)
f2 = 2*f1
Therefore, compared to the fundamental frequency the frequency would be double.
Option A is correct - While a guitar string is vibrating, you gently touch the midpoint of the string to ensure that the string does not vibrate at that point. The lowest-frequency standing wave that could be present on the string vibrates at twice the fundamental frequency.
Hi
I think the answer is:
GRAVITATIONAL POTENTIAL ENERGY TRANSFORMS INTO KINETIC ENERGY.
HOPE IT HELPS.
The total force applied by the three people is:

This force is pushing toward the direction of the motion, while the frictional force

points in the opposite direction.
We can write Newton's second law applied to the block: the resultant of the two forces must be equal to the product between the block's mass and its acceleration

We know the mass of the block, m=500 kg, and the acceleration,

, so we can find the friction:
Answer:
Explanation:
Intensity of sound = sound energy emitted by source / 4 π d² , where d is distance of the source .
A )
Intensity of sound at 1 m distance = 60 /4 π d²
d = 1 m
Intensity of sound at 1 m distance = 60 /(4 π 1²)
= 4.78 W m⁻² s⁻¹
B )
Intensity of sound at 1.5 m distance = 60 /4 π d²
d = 1.5 m
Intensity of sound at 1 m distance = 60 /(4 π 1.5²)
= 2.12 W m⁻² s⁻¹
C )
Intensity of sound due to 4 speakers at 1.5 m distance = 4 x 60 /4 π d²
d = 1.5 m
= 4 x 60 /(4 π 1.5²)
= 8.48 W m⁻² s⁻¹
D )
Intensity of sound due to .06 W speaker must be 10⁻¹² W s ⁻² . Let the distance be d .
.06 /4 π d² = 10⁻¹²
d² = .06 /4 π 10⁻¹²
d = 6.9 x 10⁴ m .