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4 years ago

Which chart correctly describes the properties of electric and magnetic fields?

Physics

1 answer:

Serjik [45]4 years ago

7 0

<span>Both electric and magnetic fields exert body forces, meaning they act from a distance. The like charges and poles in both repel; positive charge repels positive and the north pole repels the north pole. For both, the opposite poles/charges attract. Finally, only magnetic fields have poles, and there are two poles, namely the south and north, so they are dipolar. The diagram that represents all of this information correctly is the third.</span>

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3 years ago

You have 10 ohm and a 100 ohm resistor in parallel. You place this equivalent resistance in series with an LED, which is rated t

Nataly [62]

Answer:

Approximately $\rm 2.0\; V$ .

Approximately $\rm 30 \; mA$ . (assumption: the LED here is an Ohmic resistor.)

Explanation:

The two resistors here $R_1= 10\; \Omega$ and $R_2= 100\; \Omega$ are connected in parallel. Their effective resistance would be equal to

$\displaystyle \frac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2}} = \frac{1}{\dfrac{1}{10} + \dfrac{1}{100}} = \frac{10}{11} \; \Omega$ .

The current in a serial circuit is supposed to be the same everywhere. In this case, the current through the LED should be $20\; \rm mA = 0.020\; \rm A$ . That should also be the current through the effective $\displaystyle \rm \frac{10}{11} \; \Omega$ resistor. Make sure all values are in standard units. The voltage drop across that resistor would be

$V = I \cdot R = 0.020 \times \dfrac{10}{11} \approx 0.182\; \rm V$ .

The voltage drop across the entire circuit would equal to

the voltage drop across the resistors, plus
the voltage drop across the LED.

In this case, that value would be equal to $1.83 + 0.182 \approx 2.0\; \rm V$ . That's the voltage that needs to be supplied to the circuit to achieve a current of $20\; \rm mA$ through the LED.

Assuming that the LED is an Ohmic resistor. In other words, assume that its resistance is the same for all currents. Calculate its resistance:

$\displaystyle R(\text{LED}) = \frac{V(\text{LED})}{I(\text{LED})}= \frac{1.83}{0.020} \approx 91.5\; \Omega$ .

The resistance of a serial circuit is equal to the resistance of its parts. In this case,

$\displaystyle R = R(\text{LED}) + R(\text{Resistors}) = 91.5 + \frac{10}{11} \approx 100\; \Omega$ .

Again, the current in a serial circuit is the same in all appliances.

$\displaystyle I = \frac{V}{R} = \frac{3}{100} \approx 0.030\; \rm A = 30\; mA$ .

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3 years ago

8. A 30-kg box is sliding down a frictionless plane that is sloped at 24º. Assuming the object starts at rest,

Katena32 [7]

The net force on the box parallel to the plane is

∑ F[para] = mg sin(24°) = ma

where mg is the weight of the box, so mg sin(24°) is the magnitude of the component of its weight acting parallel to the surface, and a is the box's acceleration.

Solve for a :

g sin(24°) = a ≈ 3.99 m/s²

The box starts at rest, so after 7.0 s it attains a speed of

(3.99 m/s²) (7.0 s) ≈ 28 m/s

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2 years ago