Recent content by sunnnystrong

I would think not though as they gave you the length of the rods ?

Honestly, I have no idea. I am confused about how to set this problem up but found this equation in my book.

Homework Statement Two thin glass rods are placed side by side 4.70 cm apart as shown in the diagram below. They are each 11.3 cm long and are uniformly charged to +10.4 nC . Answer the questions below regarding the electric field along a line which is drawn through the middle of the two glass...

ohhh! thank you so much! the direction of the field will be 9.32 degrees above the horizontal

& The magnitude of the field would be 1.2*10^4 N/C & the direction would be 8.95 degrees above the horizontal?

Okay so I am a little confused but for Q1 the electric field would be: 10788 N/C... for Q2 it would be 2157.73 N/C... The components of Q1 would be (10788i, 0j) & for Q2 it would be (2157.7cos(60) i, 2157.7sin(60) j) ?

^^ see reply. I am a little confused about the directions.

Oh, okay so for the E1 (Top charge): E1 = ((8.99*10^9)(3*10-9)/(.05^2) i, 0j) E2 (Bottom charge): E2 = ((8.99*10^9)(3*10-9)/(.05^2) i, (8.99*10^9)(3*10-9)/(.1^2) j) Etotal = (21576 i, 2697j) |Etotal| = 21743.909 N/C ? I already tried this and they said it was wrong

Homework Statement Two +3.0nC charges are shown in the diagram below which are spaced 10cm apart. What are the strength and direction of the electric field at the position indicated by the dot in the figure (Figure 1) ? What is the electric field in N/C? What is the direction of the electric...

Take the derivative with respect to t

Foil everything... Than you're just going to be using the chain rule

By conservation of momentum m₁ u = (m₁+ m₂) v v/u = m₁ /(m₁+ m₂) EQN 1 Kinetic Energy --> (1/2)(m₁+ m₂) v ² = (1/2) 0.25 m₁ u² EQN 2

Use: X-Xo = volt + 1/2AT^2 For the police car: X-Xo = 1/2(2)(T-1)^2 + (26.4)T For the car: X-Xo = (37.5)T Set them equal to each other & solve for T :) The trick is realizing that the police car has two components for distance...

*** Figured it out. all you had to do was square the constant (16/19) as well.

Anyone? D: