Answer:
a. 30 N / m
b. 9.0 N
Explanation:
Given that
Unstretched length of the spring, 
= 20.0cm = 0.2m
a) When the mass of 4.5N is hanging from the second spring, then extended length Is
= 35.0cm = 0.35m
So, the change in spring length when mass hangs is
= (0.35 - 0.20) m
= 0.15m
As spring are identical
Let us assume that the spring constant be "k", so at equilibrium
Restoring Force on spring = Block weightage
kx = W = 4.50
= 30 N / m
b) Now for the third spring, stretched the length of spring is
= 50cm = 0.5m
So, the change in spring length is
= (0.5-0.20)m
= 0.30m
At equilibrium,
Restoring Force on spring = Block weightage
Now using all mentioned and computed values in above,
= 30(0.3)
= 9.0 N
Transverse waves travel on a direction that is perpendicular to the motion of the particles (or whatever medium is waving) So the particles must be moving east to west, which is transverse to the north-south motion of the wave
Answer:
The stars are moving away from us.
Explanation:
The observed wavelengths of hydrogen transition for stars A and B (660.0 nm and 666 nm respectively) are greater than that observed in the laboratory (656.2 nm). The observed long wavelengths for the stars means that the light from the stars is red-shifted.
According to the Doppler effect, red-shifted light means that the source is moving a way from the observer; therefore, we arrive at the conclusion that the stars A and B are moving away from us.
Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
