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Elodia [21]
3 years ago
8

The potential difference between two points is 100 V. If a particle with a charge of 2 C is transported from one of these points

to the other, what is the magnitude of the work done?
Physics
2 answers:
Sunny_sXe [5.5K]3 years ago
7 0

1 volt = 1 joule/coulomb

100 volts = 100 joules/coulomb

Work = (100 J/C) x (2 C)

Work = (+ or -) 200 J

The work could be positive or negative. It depends on direction of the potential difference, direction the charge is moved, sign of the charge, and whether we mean work done by the mover or by the potential difference.

But the MAGNITUDE of the work is 200J .

Aleonysh [2.5K]3 years ago
4 0

Answer: 200 J

Explanation: In order to explain this we have consider that the work done in a electric field is given by:

Work= Q*ΔV=2*100=200J

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If a farsighted person has a near point that is 0.600 mm from the eye, what is the focal length f2f2f_2 of the contact lenses th
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Answer:

0.22mm

Explanation:

A far sighted person is a person suffering from long sightedness i.e such individual can only see far distant object clearly but not near distant object. The defect is corrected using convex lens.

Since convex lens is used, the focal (f) length of the lens is positive and the image distance (v) is also positive.

Using the lens formula,

1/f = 1/u + 1/v

Where u is the object distance = 0.35mm

v = 0.6mm

1/f = 1/0.35+1/0.6

1/f = 2.86 + 1.67

1/f = 4.53

f = 1/4.53

f = 0.22mm

The focal length of the contact lenses will be 0.22mm

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3 years ago
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solniwko [45]
The answer would be 48
7 0
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Give three examples from your life of magnetic force.
Margaret [11]
Headphones, refrigerator magnets, and compasses

Hope that was helpful.
4 0
3 years ago
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Mention the name of the following frequencies depending upon their values a) 12Hz b) 20100Hz
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It can be a) 12Hz.................
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3 years ago
Two long parallel wires are separated by forty centimeters and carry oppositely-directed currents of ten amperes. Find the magni
Softa [21]

Answer:

1.04μT

Explanation:

Due to both wires have opposite currents, the magnitude of the total magnetic field is given by

B_T=\frac{\mu_o I}{2 \pi r_1}-\frac{\mu_o I}{2 \pi r_2}

I: electric current = 10A

mu_o: magnetic permeability of vacuum = 4pi*10^{-7} N/A^2

r1: distance from wire 1 to the point in which B is measured.

r2: distance from wire 2.

The distance between wires is 40cm = 0.4m. Hence, r1=0.2m r2=0.6m

By replacing in the formula you obtain:

B_T=\frac{(4\pi *10^{-7}N/A^2)(10A)}{2\pi}(\frac{1}{0.4m}-\frac{1}{0.6m})=1.04*10^{-6}T =1.04\mu T

hence, the magnitude of the magnetic field is 1.04μT

4 0
3 years ago
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