What is the velocity of an object that has been in free fall for 2.5 s?
1 answer:
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It is a false statement i.e. drift velocity is not same in the direction as the applied force.
Drift velocity of a current-carrying conductor can be explained as, the charges i.e. electrons do not flow in the same direction of current. In other word, in most cases the movement of the electrons is almost random, with a small net velocity. So that , the drift velocity, in the direction opposite to the electric field.
Drift velocity is inversely proportional to the number of electron per unit volume of the conductor e. Therefore, the formulation can be given as ,
= σ E/ne
The above equation shows the drift velocity in a current carrying conductor
where, is drift velocity , σ is the conductivity, E is electric force and n is number of electrons per unit volume of the conductor e.
Hence here we can say that, the drift velocity is not in the same direction as the applied force.
To know more about drift velocity
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Answer:
1). Average speed = 1.5 m per second
2). Average velocity = 1.5 m per second
Explanation:
1). Since, speed is a scalar quantity
Therefore, average speed of the trip =
From the graph attached,
Total distance covered = 10 + 10 + 20 + 0 + 20 + 30
= 90 meters
Total time taken = 60 seconds
Average speed =
= 1.5 meter per second
2). Velocity is a vector quantity.
Therefore, average velocity =
=
=
= 1.5 meter per second
Answer:
(a). The potential on the negative plate is 42.32 V.
(b). The equivalent capacitance of the two capacitors is 0.69 μF.
Explanation:
Given that,
Charge = 10.1 μC
Capacitor C₁ = 1.10 μF
Capacitor C₂ = 1.92 μF
Capacitor C₃ = 1.10 μF
Potential V₁ = 51.5 V
Let V₁ and V₂ be the potentials on the two plates of the capacitor.
(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor
Using formula of potential difference
Put the value into the formula
The potential on the second plate
(b). We need to calculate the equivalent capacitance of the two capacitors
Using formula of equivalent capacitance
Put the value into the formula
Hence, (a). The potential on the negative plate is 42.32 V.
(b). The equivalent capacitance of the two capacitors is 0.69 μF.