{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# HW 09 - saldana(avs387 homework 09 Turner(56705 This...

This preview shows pages 1–3. Sign up to view the full content.

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

View Full Document
saldana (avs387) – homework 09 – Turner – (56705) 2 = q parenleftbigg k q 2 a + 2 k q 2 a + ( 3 q ) k 2 a + 6 k q 2 a parenrightbigg = 6 k q 2 2 a . The final energy is E f = 1 2 m v 2 . From energy conversation, we have E i = E f 6 k q 2 2 a = 1 2 m v 2 v = q radicalBigg 6 2 k m a . 003 10.0 points A charged particle is connected to a string that is is tied to the pivot point P . The particle, string, and pivot point all lie on a horizontal table (consequently the figure below is viewed from above the table). The particle is initially released from rest when the string makes an angle 60 with a uniform electric field in the horizontal plane (shown in the figure). 60 1 m 209 V / m P 0 . 011 kg 3 μ C initial release ω parallel P Determine the speed of the particle when the string is parallel to the electric field.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 9

HW 09 - saldana(avs387 homework 09 Turner(56705 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online