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3 years ago

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Exit ticket: A lamp is plugged into a 110 Volt electrical outlet. There is a 9 Watt LED bulb in the lamp. a. What is

1 answer:

notka56 [123]3 years ago

5 0

Answer:

(a) 0.081 A (b) 1358.02 ohms

Explanation:

Voltage, V = 110 volt

Power of a LED bulb, P = 9 Watt

(a) Let the current is I. The formula for ower in terms of voltage and power is given by :

P = VI

$I=\dfrac{P}{V}\\\\I=\dfrac{9}{110}\\\\=0.081\ A$

(b) Let R is the resistance of the bulb. Using Ohm's law as follows :

V = IR

$R=\dfrac{V}{I}\\\\R=\dfrac{110}{0.081}\\\\=1358.02\ \Omega$

Hence, this is the required solution.

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Explanation:

(a)

From the given information:

The initial velocity $v_1$ = 5 m/s

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$v_{1 \ x } = 5 \ cos \ 30^0$

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The component along the y-axis = $v_2 { \ sin \ \theta}$

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To find the final velocity( reflected velocity)

using the same magnitude $v_2 = 5 \ m/s$

The angle from the x-axis can be $\theta_r = 90^0+60^0$

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The component along the y-axis = $v_2 \ sin \theta_r$

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(b)

The velocity $v_1$ can be written as in vector form.

$v_1 ^{\to} = v_1 x \hat {i} + v_1 y \hat {j}$

$v_1 ^{\to} =4.33 \ \hat {i} + 2.5 \ \hat {j}$ ---- (1)

The reflected velocity in vector form can be computed as:

$v_2 ^{\to} = v_2 x \hat {i} + v_2 y \hat {j}$

$v_2 ^{\to} =-4.33 \ \hat {i} + 2.5 \ \hat {j}$ --- (2)

The change in velocity = $v_2 ^{\to} - v_1 ^{\to}$

$\Delta v ^{\to} = - 4.33 \hat i + 2.5 \hat j - 4.33 \hat i - 2.5 \hat j$

$\Delta v ^{\to} = - 8.66 \hat { i }$

(c)

The magnitude of change in velocity = $| \Delta V |$

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