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Karolina [17]
2 years ago
13

If the weight of the ruler is one Newton ,Gc cannot have a value more than 25cm

Physics
1 answer:
nevsk [136]2 years ago
4 0

If the length of the ruler is 50 cm, the center of gravity cannot be greater than 25 cm.

The given parameters:

  • Weight of the ruler = 1 N

<h3>What is center of gravity (CG)?</h3>
  • Center of gravity is the point at which the weight of an object is concentrated.

Let the length of the ruler = L

The center of the gravity of the ruler is calculated as follows;

X_{CG} = \frac{W(L_0) + W(L -X_{CG})}{W} \\\\X_{CG} = \frac{1(0) + 1(L -X_{CG})}{1}\\\\X_{CG} = L - X_{CG}\\\\X_{CG } + X_{CG} = L\\\\2X_{CG} = L\\\\X_{CG} = \frac{L}{2} \\\\when , \ L = 50 \ cm\\\\X_{CG} = \frac{50}{2} \\\\X_{CG} = 25 \ cm

Thus, if the length of the ruler is 50 cm, the center of gravity cannot be greater than 25 cm. This may change if the length of the ruler changes because the center of gravity of uniform ruler depends on the length of the ruler.

Learn more about center of gravity here: brainly.com/question/6765179

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A solid conducting sphere of radius 2 cm has a charge of 8microCoulomb. A conducting spherical shell of inner radius 4 cm andout
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Answer:

C) 7.35*10⁶ N/C radially outward

Explanation:

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  • So, we can write the following equation:

       E*A = \frac{Q_{enc} }{\epsilon_{0}} (1)

  • As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
  • So, the +8 μC charge of  the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
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        Q_{enc} = - 4e-6 C - (-8e-6 C) = + 4 e-6 C

  • Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:

      E = \frac{1}{4*\pi*\epsilon_{0} } *\frac{Q_{enc} }{r^{2} } = \frac{9e9 N*m2/C2*4e-6C}{(0.07m)^{2} } = 7.35e6 N/C

  • As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
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here, m is the mass of the object, and g is the acceleration of gravity.

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