Given :
A 0.50-kg mass is attached to a spring of spring constant 20 N/m along a horizontal, frictionless surface.
The object oscillates in simple harmonic motion and has a speed of 1.5 m/s at the equilibrium position.
To Find :
At what location are the kinetic energy and the potential energy the same.
Solution :
Let, at location x from the equilibrium position the kinetic energy and the potential energy the same.
So,
Hence, this is the required solution.
Answer:
Newtons (N)
Explanation:
Because the people in charge of defining SI units said so
Answer:
The answer to your question is: 245 m
Explanation:
data
vo = 70 m/s
vf = 0 m/s
a = 10 m/s²
d = ?
Formula
vf² = vo² + 2ad
d = (vf² - vo²) / 2a
Process
d = (0² - 70²) / 2(10) substitution
d= 4900 / 20 simplify
d = 245 m
Answer:
E=26779230N/C(option 5)
Explanation:
Let Lamda be charge per lenght=73*10^-6c/m
Distance r=4.9cm =0.049m
Coulomb constant k=8.98755*10^9
But electric field is 2*k*lamda/r
E=(2*8.98755*10^9*73*10^-6)/0.049
E=26779230N/C