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amm1812
3 years ago
14

A dipole of moment 0.50 e · nm is placed in a uniform electric field with a magnitude of 1.6 105 N/C. What is the magnitude of t

he torque on the dipole for each of the following situations?
(a) the dipole is aligned with the electric field
(b) the dipole is transverse to (perpendicular to) the electric field
(c) the dipole makes an angle of 26° with the direction of the electric field
(d) Defining the potential energy to be zero when the dipole is transverse to the electric field, find the potential energy of the dipole in the electric field for the orientations specified in Parts (a) and (c).
Physics
1 answer:
gtnhenbr [62]3 years ago
5 0
Too much to read bro bro
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On Earth, 1 kg = 9.8 N = 2.2 lbs. On the Moon, 1 kg = 1.6 N = 0.37 lbs. Use these relationships to answer the following question
romanna [79]

Answer:

(a) 490 N on earth

(b) 80 N on earth

(c) 45.4545 kg on earth

(d) 270.27 kg on moon

Explanation:

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And 1 kg = 1.6 N = 0.37 lbs on moon

(a) We have given mass of the person m = 50 kg

As it is given that 1 kg = 9.8 N

So 50 kg = 50×9.8 =490 N

(b) Mass of the person on moon = 50 kg

As it is given that on moon 1 kg = 1.6 N

So 50 kg = 50×1.6 = 80 N

(c) We have given that weight of the person on the earth = 100 lbs

As it is given that 1 kg = 2.2 lbs on earth

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(d) We have given weight of the person on moon = 100 lbs

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8 0
3 years ago
If the magnitude of the magnetic field is 2.50 mT at a distance of 12.6 cm from a long straight current carrying wire, what is t
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Answer:

The magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

Explanation:

Given;

first magnetic field at first distance, B₁ = 2.50 mT

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Second magnetic field at Second distance, B₂ = ?

Second distance, r₂ = ?

Magnetic field for a straight wire is given as;

B = \frac{\mu I}{2 \pi r}

Where:

μ is permeability

B is magnetic field

I is current flowing in the wire

r distance to the wire

Let \ \frac{\mu I}{2\pi}  \ be \ constant; = K\\\\B = \frac{K}{r} \\\\K = Br\\\\B_1r_1 = B_2r_2\\\\B_2 =\frac{B_1r_1}{r_2} \\\\B_2 = \frac{2.5*10^{-3} *0.126}{0.198} \\\\B_2 = 1.591 *10^{-3}\ T\\\\B_2 = 1.591 \ mT

Therefore, the magnetic field at a distance of 19.8 cm from the wire is 1.591 mT

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