The speed of the mass v = 0.884 m/s.
<u>Explanation</u>:
Let
K1 represents the kinetic energy of the mass when it is released,
U1 represents the potential energy of the spring when the mass is released,
K2 represents the kinetic energy of the mass when the spring returns to relaxed length,
U2 represents the potential energy of the spring when the spring returns to relaxed length
The spring is stretched by 0.27 - 0.12 = 0.15 m
K1 = 0
U1 = (1/2) 
0.8 (0.15)^2
= 0.009 J
U2 = 0
By conservation of energy,
K2 + U2 = K1 + U1
K2 + 0 = 0 + 0.009 J
K2 = 0.009 J
Let v = speed of the mass
K2 = 1/2 m v^2
m = 23 g = 0.023 kg
0.009 = 1/2 0.023 v^2
0.009 = 0.0115 v^2
v = √(0.009 / 0.0115)
v = 0.884 m/s.
Answer:
a
Explanation:
it increases with increasing height
The velocity vector will move you 200 miles east in 4 hours traveling at a constant speed is 50 mph. Below is the solution:
Given:
200 miles
4 hours
200 miles/ 4 hours = 50 mph, east
Hope it helps.
Answer:
w = 12.5 rad/s and A = 50 10⁻² m
Explanation:
Let's treat this problem as a case of oscillatory movement
a) The angular velocity is given by
w² = k / m
w = √ 1.4 10⁴/90.0
w = 12.5 rad / s
b) It is climber falls by two meters, its mechanical energy at the highest point is
Em = U = mgh
Em = 90 9.8 2
Em = 1764 J
The energy of the harmonic oscillator can be calculated
E = ½ k A²
Where A is the amplitude of the movement. This is going to be the stretch for the mechanical energy of the fall
A² = 2E / k
A = √(2 1764 /1.40 10⁴)
A = 50 10⁻² m
c) To realize this part, you must know how the force constant changes with the length, in the case of two springs joined along its axis
1 /k equivalent = 1 / k1 + 1 / k2
