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Dmitry_Shevchenko [17]
3 years ago
8

A classroom has 24 fluorescent bulbs, each of which is 32 W. how much energy does it take to light the room for a minute?(unit=J

)PLEASE HELP
Physics
2 answers:
cestrela7 [59]3 years ago
3 0

Answer:

Energy= 46.08KJ

Explanation:

Given that the power needed to light each bulb is 32W

We know that Power = \frac{energy}{time}

The energy needed to light one bulb=power*time

Given time = 1minute = 60 seconds

Energy = 32W*60sec=1920J

Therefore energy needed to light one bulb is 1920J

The energy needed to light 24 bulbs = 1920*24 =46080J=46.08KJ

Alborosie3 years ago
3 0
<h2>46.08kJ</h2>

Explanation:

        There are a total of 24 bulbs in the room, each of which has a rated wattage of 32 W.

        We need to light the room for a minute. Total power demanded by bulbs=(\textrm{Number of bulbs})\times(\textrm{Wattage of each bulb})\textrm{ = }24\times32W\textrm{ = }768W.

        We now know the demanded power. Power is energy required per unit time.

Power\textrm{ = }\frac{Energy}{time}

Energy\textrm{ = }Power\times time \textrm{ = }768\textrm{ }W\times1\textrm{ }min\textrm{ = }768\textrm{ }\frac{J}{sec}\times 60\textrm{ }sec\textrm{ = }46,080J

∴ Energy required = 46,080J=46.08kJ

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Torque about the support point is given by (torque is conserved)

mgl_1=Mgl_2\\\Rightarrow M=\dfrac{ml_1}{l_2}\\\Rightarrow M=\dfrac{50\times (61.3-90)}{50-61.3}\\\Rightarrow M=126.99115\ g

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jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh
Greeley [361]

Answer:

C. 590 mph

\vert v_{cj}\vert=589.49\ mph

Explanation:

Given:

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  • velocity of Cessna, v_c=150\ mph
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Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

\vec v_j=500\ \hat i\ mph

&

\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)

\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph

Now the vector of relative velocity of Cessna with respect to jet:

\vec v_{cj}=\vec v_j-\vec v_c

\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )

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Now the magnitude of this velocity:

\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}

\vert v_{cj}\vert=589.49\ mph is the relative velocity of Cessna with respect to the jet.

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