For the answer to the question above,
<span>There is nothing in the equations to suggest that the string moves in the x direction so D) v_x(x,t)=0.
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y(x,t) = A sin(kx-omega t)
d{y(x,t)}/d{x} = A k cos(kx - omega t)
Acceleration = (change in speed) / (time for the change)
Change in speed = (end speed) - (start speed) = (15 m/s - 7 m/s) = 8 m/s
time for the change = 2 minutes = 120 seconds
Acceleration = (8 m/s) / (120 seconds)
Acceleration = 0.067 m/s²
Answer:
24m/s²
Explanation:
Given
Distance S = 3m
Time of fall = 0.5sec
Required
Acceleration due to gravity
Using the equation of motion
S = ut+1/2gt²
Substitute the given values
3 = 0+1/2g(0.5)²
3 = 1/2(0.25)g
3 = 0.125g
g = 3/0.125
g = 24
Hence the value for the acceleration of gravity on this new planet is 24m/s²
6 is the answer I remember the answer from when I took this and it was easy
