Answer:
E. The period of oscillation increases.
Explanation:
The period of oscillation is:
T = 2π√(m/k)
Frequency is the inverse of period (f = 1/T), so as period increases, frequency decreases.
Increasing the mass will increase the period and decrease the frequency.
 
        
             
        
        
        
Answer:
x = 0.974 L
Explanation:
given,
length of inclination of log = 30°
mass of log = 200 Kg
rock is located at = 0.6 L
L is the length of the log
mass of engineer = 53.5 Kg
let x be the distance from left at which log is horizontal.
For log to be horizontal system should be in equilibrium
  ∑ M = 0 
 mass of the log will be concentrated at the center  
distance of rock from CM of log = 0.1 L
now,
∑ M = 0



        x = 0.974 L
hence, distance of the engineer from the left side is equal to x = 0.974 L
 
        
             
        
        
        
Answer:
period of oscillations is 0.695 second
Explanation:
given data 
mass m = 0.350 kg
spring stretches x = 12 cm = 0.12 m
to find out
period of oscillations
solution
we know here that force 
force = k × x   .........1
so force = mg =  0.35 (9.8)  = 3.43 N
3.43 = k × 0.12
k = 28.58 N/m
so period of oscillations is
period of oscillations = 2π ×  ................2
   ................2
put here value
period of oscillations = 2π ×  
  
period of oscillations = 0.6953 
so period of oscillations is 0.695 second
 
        
             
        
        
        
Answer:L=109.16 m
Explanation:
Given
initial temperature 
Final Temperature 
mass flow rate of cold fluid 
Initial Geothermal water temperature 
Let final Temperature be T
mass flow rate of geothermal water 
diameter of inner wall 

specific heat of water 
balancing energy
Heat lost by hot fluid=heat gained by cold Fluid




As heat exchanger is counter flow therefore





heat lost or gain by Fluid is equal to heat transfer in the heat exchanger
 (LMTD)
 (LMTD)



