Answer:
a)   = 928 J
, b)U = -62.7 J
, c) K = 0
, d) Y = 11.0367 m,  e)  v = 15.23 m / s
 = 928 J
, b)U = -62.7 J
, c) K = 0
, d) Y = 11.0367 m,  e)  v = 15.23 m / s  
Explanation:
To solve this exercise we will use the concepts of mechanical energy.
a) The elastic potential energy is
        = ½ k x²
 = ½ k x²
        = ½ 2900 0.80²
 = ½ 2900 0.80²
        = 928 J
 = 928 J
b) place the origin at the point of the uncompressed spring, the spider's potential energy
      U = m h and
      U = 8 9.8 (-0.80)
      U = -62.7 J
c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also
       K = ½ m v²
       K = 0
d) write the energy at two points, maximum compression and maximum height
      Em₀ = ke = ½ m x²
       = mg y
 = mg y
      Emo = 
      ½ k x² = m g y
      y = ½ k x² / m g
      y = ½ 2900 0.8² / (8 9.8)
      y = 11.8367 m
As zero was placed for the spring without stretching the height from that reference is
      Y = y- 0.80
      Y = 11.8367 -0.80
      Y = 11.0367 m
Bonus
Energy for maximum compression and uncompressed spring
      Emo = ½ k x² = 928 J
       = ½ m v²
= ½ m v²
      Emo =  
      Emo = ½ m v²
       v =√ 2Emo / m
      v = √ (2 928/8)
      v = 15.23 m / s
 
        
             
        
        
        
Actually it's 

 and it says that the energy of an object (E) is equal to the mass (m) of the object multiplied with the squared speed of light (

). This theory says that mass can be turned into energy and energy can be turned into mass. This is one of Einstein's theory of relativity.
 
        
        
        
Answer:
The wave speed is calculated below:
Explanation:
Given,
number of waves passed per minute = 8
time period = 1 minute = 60 s
distance between successive wave crests = 20 m
waves passing interval per second =  
 
Now, 
 wave speed = 20 m ×  
 
                      =  m/s
 m/s
                      = 2.67 m/s
Hence the wave speed is 2.67 m/s.
 
        
             
        
        
        
Explanation:
Given data:
d = 30 mm = 0.03 m
L = 1m 
S = 70 Mpa
 = 70 Mpa
Δd = -0.0001d
Axial force = ?
validity of elastic deformation assumption.
Solution:
O'₂ = Δd/d = (-0.0001d)/d = -0.0001
For copper,
v = 0.326      E = 119×10³ Mpa
O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶
∵δ = F.L/E.A    and σ = F/A so,
σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa
F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN
S = 70 MPa > σ = 36.5 MPa
 = 70 MPa > σ = 36.5 MPa
∵ elastic deformation assumption is valid.
so the answer is
F = 25800 K N            and     S > σ
 > σ