The electric field strength of a point charge is inversely proportional to the square of the distance from the charge ... a lot like gravity.
If the magnitude of the field is (2E) at the distance 'd', then at the distance '2d', it'll be (2E)/(2²). That's (2E)/4 = 0.5E .
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They can be described as small in quantity and very dangerously radioactive.
Vf =Vi + at
Vi =Vf - at
= 34 - 15•13
= - 161 m/s :/
Answer: 12) 1.07 m/s (right) 13) 4.05 m/s 14) 73 m/s 15) 10.9 m/s
Explanation:
12) Conservation of momentum. Momentum is the produce of mass and velocity.
13(2) + 15(-5) = 13(-5) + 15v
v = 1.06666... ≈ 1.07 m/s (right)
13) 18(9) + 22(0) = 18v + 22v
v = 18(9)/40 = 4.05 m/s
14) 0.65(35) + 0.08(0) = 0.65(26) + 0.08v
v = 73.125
15) This is a bit trickier. Let's ASSUME you jump off at 7 m/s relative to the truck. Doing this, we can assume that the reference frame is moving along with the truck at 10 m/s
the conservation of momentum equation becomes
600(0) + 80(0) = 600v + 80(-7)
v = 0.9333333... m/s
adding back the velocity of the reference frame means the truck is now traveling.
10.9333333... ≈ 10.9 m/s
