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kotegsom [21]
4 years ago
10

Question 6. B) and c)

Physics
1 answer:
maks197457 [2]4 years ago
4 0
The formula for both is v(t) = v0 + a*t

b) v(8) = 0 + 6m/s^2 *8s = 48 m/s

now we know the beginning (2) and end speed (14), but not the time:

c) 14 = 2 + 1.5*t => t = (14-2)/1.5 = 8 seconds
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An electric dipole consisting of charges of magnitude 1.70 nC separated by 6.80 μm is in an electric field of strength 1160 N/C.
bazaltina [42]

Answer:

p = 1.16 10⁻¹⁴ C m     and  ΔU = 2.7 10 -11 J

Explanation:

The dipole moment of a dipole is the product of charges by distance

                        p = 2 a q

With 2a the distance between the charges and the magnitude of the charges

                        p = 1.7 10⁻⁹ 6.8 10⁻⁶

                        p = 1.16 10⁻¹⁴ C m

 

The potential energie dipole  is described by the expression

                       U = - p E cos θ

Where θ is the angle between the dipole and the electric field, the zero value of the potential energy is located for when the dipole is perpendicular to the electric field line

Orientation parallel to the field

                      θ = 0º

                      U = 1.16 10⁻¹⁴ 1160 cos 0

                      U1 = 1.35 10⁻¹¹ J

Antiparallel orientation

                       θ = 180º

                      cos 180 = -1

                      U2 = -1.35 10⁻¹¹ J

The difference in energy between these two configurations is the subtraction of the energies

                         ΔU = | U1 -U2 |

                         ΔU = 1.35 10-11 - (-1.35 10-11)

                         ΔU = 2.7 10 -11 J

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suppose a vacuum cleaner use 120 j of electrical energy. If 45 j are used to pull air into the vacuum cleaner, how efficient is
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