The cluster that is most likely to be located in the halo of our galaxy is the diagram that shows main-sequence stars of every spectral type except O, along with a few giants and supergiants.
<h3>What are star clusters?</h3>
Star clusters are large collections of stars. Star clusters are classified into two types: Globular clusters are gravitationally bound groups of tens of thousands to millions of old stars.
Because of their location on the dusty spiral arms of spiral galaxies, they are sometimes referred to as galactic clusters. Stars in an open cluster share a common ancestor as they all formed from the same massive molecular cloud.
A typical spiral galaxy has a faint, extended stellar halo. A stellar halo is an essentially spherical population of stars and globular clusters thought to surround most disk galaxies and the cD class of elliptical galaxies. It should be noted that a halo is a spherical cloud of stars surrounding a galaxy. Astronomers have proposed that the Milky Way's halo is composed of two populations of stars.
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Answer:
Given that
Dry-bulb temperature(T) =24°C
Wet-bulb temperature(Tw) = 17°C
Pressure ,P = 1 atm
As we know that psychrometric chart are drawn at constant pressure.
From the diagram
ω= specific humidity
Lets take these two lines Dry-bulb temperature(T) line and Wet-bulb temperature(Tw) cut at point P
From chart at point P
a)
Specific humidity,ω = 0.00922 kg/kg
b)
The enthalpy ( h)
h=47.59 KJ/kg
c)
The relative humidity, RH
RH= 49.58 %
d)
Specific volume ,
v= 0.853 m³/kg
Answer:
88.3
Explanation:
Emf in a rotating coil is given by rate of change of flux:
E= dФ/dt=(NABcos∅)/ dt
N: number of turns in the coil= 80
A: area of the coil= 0.25×0.40= 0.1
B: magnetic field strength= 1.1
Ф: angle of rotation= 90- 37= 53
dt= 0.06s
E= (80 × 0.4× 0.25×1.10 × cos53)/0.06= 88.3V
Answer:
B
Explanation:
if you sit up straight you will have a proper posture
Answer:
a = 1.72 m/s²
Explanation:
The given kinematic equation is the 2nd equation of motion. The equation is as follows:
xf = xi + (Vi)(t) + (1/2)(a)t²
where,
xf = the final position = 5000 m
xi = the initial position = 1000 m
Vi = the initial velocity = 15 m/s
t = the time taken = 60 s
a = acceleration = ?
Therefore,
5000 m = 1000 m + (15 m/s)(60 s) + (1/2)(a)(60 s)²
5000 m = 1000 m + 900 m + a(1800 s²)
5000 m = 1900 m + a(1800 s²)
5000 m - 1900 m = a(1800 s²)
a(1800 s²) = 3100 m
a = 3100 m/1800 s²
<u>a = 1.72 m/s²</u>
