The average current density in the wire is given by:
where I is the current intensity and A is the cross-sectional area of the wire.
The cross-sectional area of the wire is given by:
where r is the radius of the wire. In this problem, , so the cross-sectional area is
and the average current density is
Explanation:
(a) Displacement of an object is the shortest path covered by it.
In this problem, a student is biking to school. She travels 0.7 km north, then realizes something has fallen out of her bag. She travels 0.3 km south to retrieve her item. She then travels 0.4 mi north to arrive at school.
0.4 miles = 0.64 km
displacement = 0.7-0.3+0.64 = 1.04 km
(b) Average velocity = total displacement/total time
t = 15 min = 0.25 hour
Hence, this is the required solution.
Answer:
6000
1.2 J
Explanation:
I = Current = 1 A
t = Time = 2 ms
n = Number of electrocyte
V = Voltage = 100 mV
Charge is given by
The charge flowing through the electrocytes in that amount of time is
The maximum potential is given by
The number of electrolytes is 6000
Energy is given by
The energy released when the electric eel delivers a shock is 1.2 J
Equivalent capacitance is given by
The equivalent capacitance of all the electrocyte cells in the electric eel is
Answer:
The options are approximations of the exact answers:
A)
B)
C)
D) Toward the inner wall
E)
Explanation:
A) The electric field in a parallel plate capacitor is given by the formula , where
and in our case
and, for air,
, so we have:
B) The K+ ion has one elemental charge excess, so its charge is , and the force a charge experiments under an electric field E is given by F=qE, so we have:
C) The potential difference between two points separated a distance d under an uniform electric potential E is given by , so we have:
D) The electic field goes from positive to negative charges, so it goes towards the inner wall.
E) The work done by an electric field through a potential difference on a charge Q is
, and is equal to the kinetic energy imparted on it, so we have: