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1 year ago

When a 72 W light bulb is connected to a 298 V source, what is the current ?

Physics

1 answer:

Norma-Jean [14]1 year ago

6 0

the current is 0.24 Amperes

Explanation

power is the amount of energy transferred or converted per unit time, it can be found using the formula

$\begin{gathered} P=I*V \\ where\text{ P is the power} \\ I\text{ is the current} \\ V\text{ is the voltage} \end{gathered}$

Step 1

a)let

$\begin{gathered} P=\text{ 72 Watts} \\ V=298\text{ V} \end{gathered}$

b) now, replace in the formula and solve for I

$\begin{gathered} P=I*V \\ 72\text{ W=I*298 V} \\ divide\text{ both sides by 298 V} \\ \frac{72W}{298\text{ V}}\text{=}\frac{I*298V}{298\text{ V}} \\ 0.24\text{ Amperes=I} \end{gathered}$

therefore,the current is 0.24 Amperes

I hope this helps you

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3 years ago

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Answer:

a. $t_1=12.5\ s$

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Explanation:

Given:

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speed of lagging car, $u_{2}=35\ m.s^{-1}$

distance between the cars, $\Delta s=45\ m$

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where:

$v_1=0$ , final velocity after braking

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$v_2^2=u_2^2+2.a_2.\Delta s$

where:

$v_2=$ final velocity of the chasing car after braking = 0

$a_2=$ acceleration of the chasing car after braking

$0^2=35^2+2\times a_2\times 45$

$a_2=-13.61\ m.s^{-2}$ must be the minimum magnitude of deceleration to avoid hitting the leading car before stopping

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Answer:

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Explanation:

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