Practice Test 3

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course Phy 122

4/12 6:35 PM This is a practice test for Test 3

Thank you for your feedback." "Principles of Physics (Phy 122) Test Sets 51-53

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Problem Number 1

A circuit has a source that creates a constant 9 Volt potential difference across series resistances of 46 Ohms and 21 Ohms. What is the current through the source, and how much does the voltage change across the first, and across the second, resistor?

Total resistance is 46 Ohms + 21 Ohms = 67 Ohms

Current = V/R

Current = 9V/67Ohms = .134 amps or .134 Volts/Ohms

First resistor : .134 Volts/Ohm * 46 Ohms = 6.164 Volts

Second resistor .134 Volts/Ohm * 21 Ohms = 2.814 Volts

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Problem Number 2

What are the magnitude and the direction (in degrees) of the electric field at the point (-9.001 m,-19.01 m), due to a charge of 1 `microC at (-11.001 m,-23.01 m)?

.If we assume a test charge of 1 Coloumb, both charges are postitve and so there for repel. This will be away from (-11.001 m,-23.01 m) and towards (-9.001 m,-19.01 m),

`dx = 2m

`dy = 4m

distance = `sqrt(2m^2 + 4m^2) = 4.47m

(9 x 10^9 N m^2/C^2)( 1 x 10^-6 C)(1 C)/( 4.47 m)^2 = 450 N.

x and y components repressent 2/4.47 and 4/4.47 of this force

x component = 2/4.47 * 450 N = 201N/C

y component = 4/4.47 * 450N = 403N/C

Magnitude = `sqrt(201N/C)^2 + (403N/C)^2) = 450 N/C

Angle = arctan(4m/2m) = 63.4 deg. both are postive so it's in the first quadrant.

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Problem Number 3

A charge of 1 Coulombs moves through a displacement of .136 m, parallel to and in the direction of the field, in an electric field of 75000 N/C. How much work is done by the charge?

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1 C * .136 m * 75000N/C = 10,200 Nm which is 10,200 Joules

if we assume the direction of the field to be positive, since the force is done opposite this direction it is negative.

So the force in Newtons on this would be negative, making the Joules negative as well.

Work done by the charge would be -10,200 Joules

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Problem Number 4

Two uniform wires are made of identical materials and have identical lengths. One wire has cross-sectional diameter 2.7 mm, the other diameter 3.2 mm. At a certain potential difference the first wire carries a current of 1.1 amps. At the same potential difference and the same temperature, how much current will the second wire carry?

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.Ratios of areas will be (diamter 2/diameter 1)^2

(3.2mm/2.7mm)^2 = 1.40

The second wire will carry 1.40 * 1.1 amps = 1.54 amps

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Problem Number 5

A charge of .6 Coulombs requires 1.4 seconds to move a distance .115 m parallel to and in the direction of an electric field of 90000 N/C. How much power is required

.6 C/1.4secs * .115m * 90000N/C = 4440 Nm/s or 4440 J/s which is 4440 watts (approx)

if we assume the direction of the field to be positive, since the force is done opposite this direction it is negative.

So the force in Newtons on this would be negative, making the Joules negative as well.

This would ultimately give us negative watts or -4440watts (approx)

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Very good, and excellent attention to detail. You even reasoned out the signs correctly, which is a frequent source of error.

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