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bezimeni [28]
3 years ago
9

A copper wire has a circular cross section with a radius of 1.25 mm. If the wire carries a current of 3.70 A, find the drift spe

ed in mm/s of the electrons in this wire. (Note: the density of charge carriers (electrons) in a copper wire is n = 8.46 × 1028 electrons/m3)
Physics
1 answer:
yulyashka [42]3 years ago
8 0

Answer:

0.0557 mm / s

Explanation:

r = 1.25 mm = 1.25 x 10^-3 m, i = 3.7 A, n = 8.46 x 10^28 per cubic metre,

e = 1.6 x 10^-19 C

Let vd be the drift velocity

use the formula for the current in terms of drift velocity

i = n e A vd

3.7 = 8.46 x 10^28 x 1.6 x 10^-19 x 3.14 x 1.25 x 1.25 x 10^-6 x vd

vd = 5.57 x 10^-5 m/s

vd = 0.0557 mm / s

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A small truck has a mass of 2145 kg. How much work is required to decrease the speed of the vehicle from 25.0 m/s to 12.0 m/s on
MAXImum [283]

Answer:

The work required is -515,872.5 J

Explanation:

Work is defined in physics as the force that is applied to a body to move it from one point to another.

The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.

Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:

Ec=\frac{1}{2} *m*v^{2}

where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).

The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:

W=\frac{1}{2} *m*v2^{2}-\frac{1}{2} *m*v1^{2}

W=\frac{1}{2} *m*(v2^{2}-v1^{2})

In this case:

  • W=?
  • m= 2,145 kg
  • v2= 12 \frac{m}{s}
  • v1= 25 \frac{m}{s}

Replacing:

W=\frac{1}{2} *2145 kg*((12\frac{m}{s} )^{2}-(25\frac{m}{s} )^{2})

W= -515,872.5 J

<u><em>The work required is -515,872.5 J</em></u>

3 0
3 years ago
a steel block has a mass of 40g.it is in the form of a cube. each edge is 1.74cm long. calculate the density
Vinil7 [7]

Answer:

d ≈ 7,6 g/cm³  

Explanation:

d = m/V = 40g/5,27cm³ ≈ 7,6 g/cm³

V = l³ = (1.74cm)³ ≈ 5,27 cm³

3 0
3 years ago
I need help on all of this
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4 0
3 years ago
According to the graph, how many atoms would remain after two half-lives?
Fudgin [204]

Answer:

Let No be initial no of atoms

N = N0 / 2      after 1 half-life

N = N0 / 4     after 2 half-lives

So after 2 half-lives 20 of the 80 atoms remain

4 0
2 years ago
Define an element and give 5 examples of elements that are important to life .
fenix001 [56]

Answer:

A pure substance consisting only of atoms with the same number of protons in their nuclei-these appear on the periodic table

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Carbon

Sulfur

Phosphate

Nitrogen

Magnesium

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Potassium

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8 0
2 years ago
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