Complete question:
A 200 g load attached to a horizontal spring moves in simple harmonic motion with a period of 0.410 s. The total mechanical energy of the spring–load system is 2.00 J. Find
(a) the force constant of the spring and (b) the amplitude of the motion.
Answer:
(a) the force constant of the spring = 47 N/m
(b) the amplitude of the motion = 0.292 m
Explanation:
Given;
mass of the spring, m = 200g = 0.2 kg
period of oscillation, T = 0.410 s
total mechanical energy of the spring, E = 2 J
The angular speed is calculated as follows;
(a) the force constant of the spring
(b) the amplitude of the motion
E = ¹/₂kA²
2E = kA²
A² = 2E/k
Answer:
a. wavelength of the sound, 
b. observed frequecy, 
Given:
speed of sound source, 
= 80 m/s
speed of sound in air or vacuum, 
= 343 m/s
speed of sound observed, 
= 0 m/s
Solution:
From the relation:
v = 
(1)
where
v = velocity of sound
= observed frequency of sound
= wavelength
(a) The wavelength of the sound between source and the listener is given by:
(2)
(b) The observed frequency is given by:
(3)
Using eqn (2) and (3):
Answer:
Time taken by A and B is 1.2 hr.
Explanation:
Given that
Time taken by tank when all(A+B+C) are open = 1 hr
Time taken by tank when A+C are open = 1.5 hr
Time taken by tank when B+C are open = 2 hr
If we treat as filling of tank is a work then
Work = time x rate
Lets take work is 1 unit
1 = 1(1/a+1/b+1/c) ---------1
1 = 1.5(1/a+1/c) ----------2
1 = 2(1/b+1/c) --------3
From equation 1 and 3
1=1(1/a+1/2)
a=2
Form equation 2
1 = 1.5(1/2+1/c)
c=6
From equation 3
1 = 2(1/b+1/6)
b=3
So time taken by
A is alone to fill tank is 2 hr
B is alone to fill tank is 3 hr
C is alone to fill tank is 6 hr
So 
Time taken by A and B is 1.2 hr.
