HW-_Fields_combo_with_other_topics

040 kg cart is moving at a speed of 6.0 m/s when it

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

View Full Document Right Arrow Icon

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

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

Unformatted text preview: 040 kg cart is moving at a speed of 6.0 m/s when it enters a 1.8 x 10 4 N/C electric field that stops the cart in 0.40 m. Determine the electric charge on the cart. 1x10-4 C (HINT: Remember these problems combine your knowledge of Electric Fields, Newton’s 2 nd Law, and our 1 st 4 math reps of Kinematics. You will have to use all of these in some order.) Since this is only the second time you’ve seen this type of problem, there’s a really big hint…. The order will be kinematics, newton’s 2 nd , then e-field. Kinematics and Newton’s 2 nd are connected by acceleration. Newton’s 2 nd and E-Field are connected by Force. 2) A charge of 4x10-6 C starts from rest in an electric field with a strength of 520 N/C. After 0.03 s, the charge is traveling at 3.5x10 5 m/s. What is the mass of the charge? 1.78x10-10 kg 3) A 2 kg mass with charge of 5x10-3 C starts from rest in an electric field with a strength of 2.4x10 5 N/C. What will its velocity be after 0.4 s? 240 m/s 4) A 0.50 kg cart with charge 3.0 x 10 –6 C slides on a horizontal surface that exerts a 2.0 N friction force on the cart. The cart moves through a 1.0 x 10 6 N/C horizontal constant r E field that points toward the right. If the cart starts at rest, determine its speed after moving 2.0 m. (HINT: remember… the sum of the forces equals mass multiplied by acceleration. 2.83 m/s 0.5m 5,000 N/C + + + + + + + + q ?...
View Full Document

{[ snackBarMessage ]}

Page1 / 3

040 kg cart is moving at a speed of 6.0 m/s when it enters...

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

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