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Dominik [7]
3 years ago
7

How are the spiral arms of the milky way detected?

Physics
1 answer:
Step2247 [10]3 years ago
6 0
The spiral structure emerges when galactic clusters (open), H II regions and O & B type stars (young stars) are used as tracers. We know this to be true as other pinwheel galaxies exhibit the same patterns across these tracers as in the milky way.
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Consider a glider flying at 400 meters altitude, when suddenly all its static ports become blocked by volcanic ash. The pressure
Furkat [3]

Answer:

the airspeed indicated by the pitot-tube driven airspeed indicator is 91.23m/s

Explanation:

Pitot tube

U = \sqrt{\frac{(p_t - p_s)2}{d} }

U = velocity(m/s)

p_t= stagnation pressure (pa)

p_s= static pressure (pa)

d = fluid density(kg/m³)

p_t = p_a_t_m + \frac{1}{2} dv^2

v = true velocity

= 101325 + 1/2(1.225)(25)²

p_t = 101,707.8125pa

p_s = 96,610pa

d = 1.225kg/m³

U = \sqrt{\frac{2(101,707.8125 - 96,610)}{1.225} } \\\\U = 91.23m/s

the airspeed indicated by the pitot-tube driven airspeed indicator is 91.23m/s

5 0
3 years ago
Which is true about a concave mirror? Incident rays that are parallel to the central axis are dispersed but will be perceived as
Reil [10]

Answer:

'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

Explanation:

The question is incomplete, find the complete question in the comment section.

Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. <em>The focus point is close to the curved  mirror than the centre of curvature.</em>

<em></em>

During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. <em>All incident light striking the surface all converges at a point on the central axis known as the focus.</em>

Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.

5 0
3 years ago
PLEASE HELPPPP !!!! Scientific investigations must be well ____ to be sure the data collected will help answer the scientists qu
konstantin123 [22]
Answer-

Analyzed or conducted

Hope this helped
3 0
2 years ago
Help me please forget about you and she was
ra1l [238]

The figure is showing a volume of 2.4 mL becuase it's feel 4 little segments.

Therefore, the answer is 2.4 mL.

5 0
1 year ago
A flywheel in a motor is spinning at 590 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75
Gnom [1K]

Answer:

Explanation:

Hello,

Let's get the data for this question before proceeding to solve the problems.

Mass of flywheel = 40kg

Speed of flywheel = 590rpm

Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.

Time = 30s = 0.5 min

During the power off, the flywheel made 230 complete revolutions.

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + ω₂) / 2] × 0.5

But ∇θ = 230 revolutions

∇θ/t = (530 + ω₂) / 2

230 / 0.5 = (530 + ω₂) / 2

Solve for ω₂

460 = 295 + 0.5ω₂

ω₂ = 330rpm

a)

ω₂ = ω₁ + αt

but α = ?

α = (ω₂ - ω₁) / t

α = (330 - 590) / 0.5

α = -260 / 0.5

α = -520rev/min

b)

ω₂ = ω₁ + αt

0 = 590 +(-520)t

520t = 590

solve for t

t = 590 / 520

t = 1.13min

60 seconds = 1min

X seconds = 1.13min

x = (60 × 1.13) / 1

x = 68seconds

∇θ = [(ω₂ + ω₁) / 2] × t

∇θ = [(590 + 0) / 2] × 1.13

∇θ = 333.35 rev/min

8 0
3 years ago
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