9 × 10²¹ electrons flow through a cross section of the wire in one hour.
<h3>What is the relation between current and charge?</h3>
- Mathematically, current = charge / time
- In S.I. unit, Charge is written in Coulomb and time in second.
<h3>What is the amount of charge flown through a wire for one hour if it carries 0.4 A current?</h3>
- Charge= current × time
- Current= 0.4 A, time = 1 hour= 3600 s
- Charge= 0.4× 3600
= 1440 C
<h3>How many numbers of electrons present in 1440C of charge?</h3>
- One electron= 1.6 × 10^(-19) C
- So, 1440 C = 1440/1.6 × 10^(-19)
= 9 × 10²¹ electrons
Thus, we can conclude that the 9 × 10²¹ electrons flow through a cross section of the wire in one hour.
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There are none on the list you included with your question.
Answer:
Minimum work = 5060 J
Explanation:
Given:
Mass of the bucket (m) = 20.0 kg
Initial speed of the bucket (u) = 0 m/s
Final speed of the bucket (v) = 4.0 m/s
Displacement of the bucket (h) = 25.0 m
Let 'W' be the work done by the worker in lifting the bucket.
So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.
Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,
Therefore, the work done by the worker in lifting the bucket is given as:
Now, plug in the values given and solve for 'W'. This gives,
Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.
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