Answer:
96%
Explanation
Let A the total area of the galaxy, is modeled as a disc:
A = πR^2 = π (25 kpc)^2
And let a be the area that astronomers are able to see:
a = πr^2 = π(5 kpc)^2
The percentage that can be seen is equal to 100 times the ratio of the areas, of the galaxy and the "visible" part:
P = 100 a/A = (5/25)^2 = 100/25 = 4%
Therefore, the percentage of the galaxy not included, i.e. not seen is:
(100-4)% = 96%
Answer:
87.5 m/s
Explanation:
The speed of a wave is given by

where
v is the wave speed
is the wavelength
f is the frequency
In this problem, we have
is the frequency
is the wavelength
Substituting into the equation, we find

A)We know the formula of the angular speed is ω = 2π / TWhere T is the time period.When second hand completes one revolution then the time taken is 60s.So T = 60sThen the angular speed of the second hand is ω= 2π / (60s) = 0.1047 rad/sb)When the minute hand completes one revolution the time taken is T = 1 hr = 3600sThen the angular speed of the minute hand is ω =(2π) / (3600s) = 0.001745 rad/sc)When the hour hand completes one revolution then the timeperiod is T = 12hrs = (12)(3600)sThen the angular speed of the hour hand is ω =(2π) / [(12)(3600)s] = 1.45444 x 10^-4 rad/s
Answer:
t = 5.05 s
Explanation:
This is a kinetic problem.
a) to solve it we must fix a reference system, let's use a fixed system on the floor where the height is 0 m
b) in this system the equations of motion are
y = v₀ t + ½ g t²
where v₀ is the initial velocity that is v₀ = 0 and g is the acceleration of gravity that always points towards the center of the Earth
e) y = 0 + ½ g t²
t = √ (2y / g)
t = √(2 125 / 9.8)
t = 5.05 s
<span>The loudness of the sound increases gradually as the air is slowly introduced in to the jar. This is because sound needs a physical medium and in a vacuum there is none. The air provides that medium and as it is introduced, the transfer of sound energy increases</span>