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Leya [2.2K]
4 years ago
7

Concept map for kinetic energy, work and power

Physics
1 answer:
Helen [10]4 years ago
8 0

Kinetic energy: the energy of motion

Work: the change in kinetic energy

Power: the rate of work done

Explanation:

The kinetic energy of an object is the energy possessed by the object due to its motion. Mathematically, it is given by:

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

The work done an object is the amount of energy transferred; according to the energy-work theorem, it is equal to the change in kinetic energy of an object:

W=K_f - K_i

where

K_f is the  final kinetic energy

K_i is the initial kinetic energy

Finally, the power is the rate of work done per unit time. Mathematically, ti can be expressed as

P=\frac{W}{t}

where

W is the work done

t is the time elapsed

Learn more about kinetic energy, work and power:

brainly.com/question/6536722

brainly.com/question/6763771

brainly.com/question/6443626

brainly.com/question/7956557

#LearnwithBrainly

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Explanation:

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andreev551 [17]

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3 0
3 years ago
The free-electron density in a copper wire is 8.5×1028 electrons/m3. The electric field in the wire is 0.0520 N/C and the temper
meriva

Answer:

(a) 1.87 x 10⁻⁴ m/s

(b) 0.013V

Explanation:

(a) Drift speed, v_{d} , is the average velocity that a charged particle can have due to an electric field. For a given current, I, the drift velocity is given by;

v_{d} = \frac{I}{qnA}             ----------------(i)

Where;

q = amount of charge

n = free charge density

A = cross-sectional area of the wire

But current density, J, is the electric current per unit cross-section area. This  is also equal to the ratio of the electric field, E, to the resistivity, p, of the material of the wire. i.e

J = \frac{I}{A} = \frac{E}{p}

Equation (i) can then be written as follows;

v_{d} = \frac{J}{qn} = \frac{E}{qnp}

v_{d} = \frac{E}{qnp}      ---------------------(ii)

From the question;

E = 0.0520N/C

p = 1.72 x 10⁻⁸ Ωm

n = 8.5 x 10²⁸ electrons/m³

c = charge on electron = 1.9 x 10⁻¹⁹C

Substitute these values into equation (ii) as follows;

v_{d} = \frac{0.0520}{1.9*10^{-19} * 8.5*10^{28} * 1.72*10^{-8}}

v_{d} = 1.87 x 10⁻⁴ m/s

(b) The potential difference, V, is given by the product of the electric field and the distance, d, between the two points in the wire. i.e

V = E x d        [where d = 25.0cm = 0.25m]

V = 0.0520 x 0.25

V = 0.013V

4 0
3 years ago
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