Double Question, 20 points and brainlyest if possible
1 answer:
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Answer:
Explanation:
Electric field strength= Force/unit charge
E= (kQq/r²)/q ₓ r
where r is the unit vector in the direction of unit charge
E=
Answer:
Three long wires are connected to a meter stick and hang down freely. Wire 1 hangs from the 50-cm mark at the center of the meter stick and carries 1.50 A of current upward. Wire 2 hangs from the 70-cm mark and carries 4.00 A of current downward. Wire 3 is to be attached to the meterstick and to carry a specific current, and we want to attach it at a location that results ineach wire experiencing no net force.
(a) Determine the position of wire 3.
b) Determine the magnitude and direction of current in wire 3
Explanation:
a)
position of wire = 50 - 1.2
= 48.8cm
b)
Direction ⇒ downward
Answer:
a) 23 m/s
Explanation:
- Assuming no external forces acting during the collision, total momentum must be conserved, as follows:
- The initial momentum p₀, can be written as follows:
- The final momentum pf, can be written as follows:
- Since (2) and (3) are equal each other, we can solve for the only unknown that remains, v₂₀, as follows:
- This means that the 2.0-kg object was moving at 23 m/s in a direction opposite to the 6.0-kg object, so its initial speed, before the collision, was 23.0 m/s.