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sesenic [268]
3 years ago
8

A 14 kilogram beam is hinged at one end. A 6.0-kilogram triangular object and a 7.5-kilogram capital I-shaped object are placed

as shown in the figure. The dots indicate the individual centers of gravity of the beam and objects. The center of mass of the entire system must be:

Physics
1 answer:
GuDViN [60]3 years ago
5 0

Answer: 1.3 m

Explanation:

From the figure attached, 3 materials are involved. A 14 kilogram beam hinged at one end, a 6.0-kilogram triangular object and a 7.5-kilogram capital I-shaped object.

Let take the moment at the hinge axis.

The moment for the beam will be

14 × ( 0.36 + 0.53 ) = 12.6

The moment of the triangle object will be 6 × 0.36 = 2.16

The moment of capital I shape will be

7.5 × ( 0.36 + 0.53 + 1.92 ) = 21.075

The total moment = 12.6 + 2.16 + 21.075

The total moment = 35.835

The total mass = 14 + 6 + 7.5 = 27.5

The center of mass of the entire system must be:

X = total moment ÷ total mass

Substitute both in the formula

X = 35.835/27.5

X = 1.3 m.

Therefore, The center of mass of the entire system must be 1.3m

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The closest distance a book can be read from a pair of reading eyeglasses (Power = 1.55 dp) is 26.0 cm. What is the near distanc
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Answer:

The image distance is 20.0 cm.

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\dfrac{1}{f}=\dfrac{1}{u}+\dfrac{1}{v}

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3 years ago
a body of mass 0.2kg is whirled round a horizontal circle by a string inclined at 30 degrees to the vertical calculate <br /&
Katen [24]

Answer:

a)  T = 2.26 N, b) v = 1.68 m / s

Explanation:

We use Newton's second law

Let's set a reference system where the x-axis is radial and the y-axis is vertical, let's decompose the tension of the string

        sin 30 = \frac{T_x}{T}

        cos 30 = \frac{T_y}{T}

        Tₓ = T sin 30

        T_y = T cos 30

Y axis  

       T_y -W = 0

       T cos 30 = mg                     (1)

X axis

        Tₓ = m a

they relate it is centripetal

        a = v² / r

we substitute

         T sin 30 = m\frac{v^2}{r}            (2)

a) we substitute in 1

         T = \frac{mg }{cos 30}

         T = \frac{ 0.2 \ 9.8}{cos  \ 30}

         T = 2.26 N

b) from equation 2

           v² = \frac{T \ sin 30 \ r}{m}

If we know the length of the string

          sin 30 = r / L

          r = L sin 30

we substitute

          v² = \frac{ T \ sin 30 \ L \ sin 30}{m}

          v² = \frac{TL \ sin^2  30}{m}

For the problem let us take L = 1 m

let's calculate

          v = \sqrt{ \frac{2.26 \ 1 \ sin^230}{0.2} }

          v = 1.68 m / s

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