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4 years ago

12

If the batteries in a portable cd player provide a terminal voltage of 12.0 v, what is the potential difference across the entir

e cd player?

1 answer:

jekas [21]4 years ago

6 0

Potential difference across the entire cd player is 12 V

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Why is your weight the same any place on Earth?

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A uniform meter rule of weight 0.9N is suspended horizontally by two vertical loops of thread A and B placed 20cm and 30cm from

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The distance from the centre of the rule at which a 2N weight must be suspend from A is 29.3 cm.

<h3>Distance from the center of the meter rule</h3>

The distance from the centre of the rule at which a 2N weight must be suspend from A is calculated as follows;

-----------------------------------------------------------------

20 A (30 - x)↓ x ↓ 20 cm B 30 cm

2N 0.9N

Let the center of the meter rule = 50 cm

take moment about the center;

2(30 - x) + 0.9(x)(30 - x) = 0.9(20)

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2 years ago

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Building codes usually limit the current carried by a No. 14 copper wire to 15 A. Many household circuits are wired with this si

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The drift velocity of the electrons in this case is c) 5.52 x 10-4 m/s

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First of all, we need to find the volume occupied by 63.3g of copper, so $V=\frac{m}{\rho}= \frac{0.0633Kg}{8,900kg/m^{3}} =7.11 \times 10^{-6} m^{3}$ , then if we assume each atom of copper contributes with one free electron to the material body $n=\frac{N_{A}}{V}=\frac{6.02 \times 10^{23} electrons}{7.11 \times 10^{-6} m^{3}} =8.46 \times 10^{28} electrons/m^{3}$ .

Finally, we apply $v_{d}=\frac{I}{nqA}$ , where A is area so, $A=\pi (\frac{d}{2})^{2}= \pi (\frac{0.0016m}{2})^{2}=2 \times 10^{-6} m^{2}$ , thus $v_{d}=\frac{I}{nqA}=\frac{15C/s}{(8.46 \times 10^{28} m^{-3})(1.60 \times 10^{-19} C)(2 \times 10^{-6} m^{2})}=5.50 \times 10^{-4} m/s$ .

As we can see c. 5.52 x 10-4 m/s is the nearest one.

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