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babymother [125]
3 years ago
7

A uniform meter ruler is balanced at its midpoint

Physics
1 answer:
NeTakaya3 years ago
4 0

Answer:

a) i) x = 0.25 m, ii) x = 0.10 m, iii)  x = 0.050 m

b) i)  x = 0.40 m

Explanation:

a) For this exercise we use the rotational equilibrium equation, where we assume that the anticlockwise rotations are positive.

1) L = 2W  

we set our reference system in the center of the bar where the fulcrum is

              ∑τ = 0

              W 0.50 - L x = 0

              x = 0.50 W / L

we substitute the value

              x = 0.50 W / 2W

              x = 0.25 m

ii) L = 5W

we calculate

              x = 0.50 W / 5W

              x = 0.10 m

iii) L = 10 W

               x = 0.50 W / 10W

              x = 0.050 m

b) a new weight is placed at x₂ = 30 cm on the left side

              W 0.50 + W 0.30 - L x = 0

              x = (0.50 + 0.30) W / L

              x = 0.80 W / L

we calculate

i)  L = 2W

                x = 0.80 w / 2w

                x = 0.40 m

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The net vertical force on the balloon is calculated as follows;

F_y = 514 \ N \ \ - \ \ 267 \ N\\\\F_y = 247 \ N

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F_x = 678 \ N \ - \ 397 \ N\\\\F_x = 281 \ N

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An object is placed in front of a convex lens of a length 10cm. What is the nature of the image formed if the object distance is
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{\mathfrak{\underline{\purple{\:\:\: Given:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Focal\:length=10\:cm}

\:\:\:\:\bullet\:\:\:\sf{Object \ distance = -15\:cm}

\\

{\mathfrak{\underline{\purple{\:\:\:To \:Find:-\:\:\:}}}} \\ \\

\:\:\:\:\bullet\:\:\:\sf{Nature \: of \:the\:image}

\\

{\mathfrak{\underline{\purple{\:\:\: Solution:-\:\:\:}}}} \\ \\

<h3>☯ <u>By using formula of Lens</u> </h3>

\\

\dashrightarrow\:\: {\boxed{\sf{\dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}}}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v}-\dfrac{1}{-15}=\dfrac{1}{10}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v}+\dfrac{1}{15}=\dfrac{1}{10}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{10} - \dfrac{1}{15}}

\\

\dashrightarrow\:\: \sf{\dfrac{1}{v} = \dfrac{1}{30}}

\\

\dashrightarrow\:\: \sf{ v = 30 \ cm}

\\

<h3>☯ <u>Now, Finding the magnification </u></h3>

\\

\dashrightarrow\:\: \sf{ m = \dfrac{-30}{-15}}

\\

\dashrightarrow\:\: \sf{m = -2}

\\

<h3>☯ <u>Hence</u>,\\</h3>

\:\:\:\:\star\:\:\:\sf{Image \ distance = 30 \ cm}

\:\:\:\:\star\:\:\:\sf{Nature = Real \ \& \ inverted}

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Happy Holidays!

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