chapter 7 - The mass of Helium = 4x1.00794 = 4.03176 The...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Astronomy –odd- numbered –Homework Chapter 7 1. Radius of the sun = 0.7 .10^6 km The chromospheres is 2500 km thick So, the sun radius = 2500/ 0.7 .10^6 = 0.0035 = 0.35% 3. 1ly = 9.46.10^12 km 5ly = 47.3x10^12 km Diameter of the sun = 1.39x10^6 km So, the angular size = (206256)(1.39.10^6 / 47.3.10^12) = 0.0061 seconds of arc The resolution of the Hubble Space telescope is about 0.023, so it would not be able to resolve detail on the surface of such a star at this distance 5. E= mc 2 The mass of hydrogen = 1.00794 ,
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Background image of page 2
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: The mass of Helium = 4x1.00794 = 4.03176 The helium atom = 4.002602, so the difference is 4.03176 – 4.002602 = 0.02914 (0.02914) / (4.03176) = 0.007 E = mc 2 = (0.007).(3.10^8)^2 = 0.063.10^16 7. L = 3.846x10^26W L = ( Δm/Δt). c 2 ( Δm/Δt) = L/ c 2 = (3.846x10^26) )/(3.10^8)^2 = 0. 427x10^10 kg/s 9. 1 megaton bomb produces about 4.10^15 J, 10^25 solar flare can release about : (10^25) / (4.10^15) = 0.25x10^10 = 2,500,000,000 of TNT...
View Full Document

{[ snackBarMessage ]}

Page1 / 2

chapter 7 - The mass of Helium = 4x1.00794 = 4.03176 The...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online