distant galaxy emits light that has a wavelength of 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm.
1 answer:
Answer:
The speed of the galaxy relative to the Earth is .
Explanation:
We have,
(a) Wavelength emitted by light at distant galaxy is 434.1 nm. On earth, the wavelength of this light is measured to be 438.6 nm. It can be seen that the wavelength of light reduces as it reaches Earth. It is called Red shift. As per Doppler's effect, we can say that the galaxy is receding from the Earth.
(b) Let v is the speed of the galaxy relative to the Earth. It can be given by :
So, the speed of the galaxy relative to the Earth is .
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1) The period of a simple pendulum depends on B) III. only (the length of the pendulum)
2) The angular acceleration is C)
3) The frequency of the oscillation is C) 1.6 Hz
4) The period of vibration is B) 0.6 s
5) The diameter of the nozzle is A) 5.0 mm
6) The force that must be applied is B) 266.7 N
Explanation:
1)
The period of a simple pendulum is given by
where
T is the period
L is the length of the pendulum
g is the acceleration of gravity
From the equation, we see that the period of the pendulum depends only on its length and on the acceleration of gravity, while there is no dependence on the mass of the pendulum or on the amplitude of oscillation. Therefore, the correct option is
B) III. only (the length of the pendulum)
2)
The angular acceleration of the rotating disc is given by the equation
where
is the final angular velocity
is the initial angular velocity
t is the time elapsed
For the compact disc in this problem we have:
(since it starts from rest)
is the final angular velocity
t = 2 s
Substituting, we find
3)
For a simple harmonic oscillator, the acceleration and the displacement of the system are related by the equation
where
a is the acceleration
x is the displacement
is the angular frequency of the system
For the oscillator in this problem, we have the following relationship
which implies that
And so
Also, the angular frequency is related to the frequency by
Therefore, the frequency of this simple harmonic oscillator is
4)
When the mass is hanging on the sping, the weight of the mass is equal to the restoring force on the spring, so we can write
where
m is the mass
is the acceleration of gravity
k is the spring constant
x = 8.0 cm = 0.08 m is the stretching of the spring
We can re-arrange the equation as
The angular frequency of the spring is given by
And therefore, its period is
5)
According to the equation of continuity, the volume flow rate must remain constant, so we can write
where
is the cross-sectional area of the hose, with
being the radius of the hose
is the speed of the petrol in the hose
is the cross-sectional area of the nozzle, with
being the radius of the nozzle
is the speed in the nozzle
Solving for , we find the radius of the nozzle:
So, the diameter of the nozzle will be
6)
According to the Pascal principle, the pressure on the two pistons is the same, so we can write
where
is the force that must be applied to the small piston
is the area of the first piston, with
being its radius
is the force applied on the bigger piston (the weight of the car)
is the area of the bigger piston, with
being its radius
Solving for , we find
So, the closest answer is B) 266.7 N.
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