Kinetic energy =(1/2) (mass) (speed²)
First object: KE = (1/2) (2 kg) (2m/s)² = 4 joules during the lift.
Second object: KE = (1/2) (4kg) (3 m/s)² = 18 joules during the lift.
The second object has more kinetic energy while it's being lifted
than the first object has while it's being lifted. Once they reach their
final heights and stop, neither object has any kinetic energy.
I think you can just sub the values in? unless the qn is asking for smth else?
Answer:
I = 4.75 A
Explanation:
To find the current in the wire you use the following relation:
(1)
E: electric field E(t)=0.0004t2−0.0001t+0.0004
ρ: resistivity of the material = 2.75×10−8 ohm-meters
J: current density
The current density is also given by:
(2)
I: current
A: cross area of the wire = π(d/2)^2
d: diameter of the wire = 0.205 cm = 0.00205 m
You replace the equation (2) into the equation (1), and you solve for the current I:

Next, you replace for all variables:

hence, the current in the wire is 4.75A