{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Becomes mgya yb paxmgx q ybcotob from which pax

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ce Ko/ ILlEI = 5.27, five complete crossings are made, leaving the block with energy Ko. - 5j..6.EI = 0.113 J on the curved rise side. This remaining energy is sufficient to move the block a distance s = 0.113 J ..;J-tkmg = 22.5 em towards point 0, so the block comes to rest 85 - 22.5 = 62.5 ern to the right of point O. g slides back and forth in a hemispherical lof 11 cm radius, starting from rest at the as shown in Fig. 8-40. The bowl is frictionless pt for a L5-cm-wide sticky patch at the tom, where the coefficient of friction is 0.6l. ,",many times does the bug cross the sticky Jf6 5""''1 FrictioDleu FIGURE (J P~~:J:1 Frictionless 8-41 Problem 61 Solution. Section 8-7: Conservation of Energy and Mass-Energy Problem 62. Two deuterium nuclei fuse to form a helium nucleus. Each deuterium has mass 3.344x 10-27 kg, and the helium has mass 6.645xlO-27 kg. Find the energy released in this reaction. FIGURE 8-40 Problem 60. Solution The difference in mass between two deuterium nuclei and one helium nucleus...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online