Answer:
The solution was to implement an absolute magnitude scale to provide a reference between stars. To do so, astronomers calculate the brightness of stars as they would appear if it were 32.6 light-years, or 10 parsecs from Earth.
Explanation:
the answer is a because they start on the north side
Answer:
22m/s
Explanation:
To find the velocity we employ the equation of free fall: v²=u²+2gh
where u is initial velocity, g is acceleration due to gravity h is the height, v is the velocity the moment it hits the ground, taking the direction towards gravity as positive.
Substituting for the values in the question we get:
v²=2×9.8m/s²×25m
v²=490m²/s²
v=22.14m/s which can be approximated to 22m/s
Answers:(a) 
μT
(b) 
μm
(c) f =
Explanation:Given electric field(in y direction) equation:

(a) The amplitude of electric field is

. Hence
The amplitude of magnetic field oscillations is

Where c = speed of light
Therefore,

μT (Where T is in seconds--signifies the oscillations)
(b) To find the wavelength use:



μm
(c) Since c = fλ
=> f = c/λ
Now plug-in the values
f = (3*10^8)/(0.4488*10^-6)
f =
Answer:
0.231 m/s
Explanation:
m = mass attached to the spring = 0.405 kg
k = spring constant of spring = 26.3 N/m
x₀ = initial position = 3.31 cm = 0.0331 m
x = final position = (0.5) x₀ = (0.5) (0.0331) = 0.01655 m
v₀ = initial speed = 0 m/s
v = final speed = ?
Using conservation of energy
Initial kinetic energy + initial spring energy = Final kinetic energy + final spring energy
(0.5) m v₀² + (0.5) k x₀² = (0.5) m v² + (0.5) k x²
m v₀² + k x₀² = m v² + k x²
(0.405) (0)² + (26.3) (0.0331)² = (0.405) v² + (26.3) (0.01655)²
v = 0.231 m/s