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Anna11 [10]
3 years ago
8

Which of the following are not examples of circular motion? Check all that apply

Physics
1 answer:
CaHeK987 [17]3 years ago
5 0

B.riding on a ferris wheel

E.a satellite orbiting the Earth

Explanation:

A person riding on a ferris wheel and a satellite orbiting the earth both depicts circular motion.

Circular motion is the motion of a body about a fixed axis or center.

  • The motion path of the body forms a circle around the center.
  • In circular motion, the travel path is along a circle.
  • An orbiting satellite traces out its orbit round the earth.
  • At certain times, it passes through that point over and over again.
  • This is similar to that of a ferris wheel

Other options gives other types of motion.

learn more:

circular motion brainly.com/question/2562955

#learnwithBrainly

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That's Newton's 1st law of motion, sometimes called the inertia law.
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Ninas measurements shown in the table here BEST represent a wave with
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They best represent a wave with zero energy and zero amplitude.

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How much farther from the charge is the 2000 v equipotential surface than the 3000 v surface?
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Answer:

r1 -r2 = 3.75cm

Explanation:

Check the attached file for the solution

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Suppose I have a spherical insulating shell with inner radius r and outer radius R. The shell has a uniformly distributed charge
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3Q / 4 pi (R^3 - r^3)

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4 0
3 years ago
A uniform horizontal bar of mass m1 and length L is supported by two identical massless strings. String A Both strings are verti
NeX [460]

Answer:

a)  T_A = \frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} ) ,  b) T_B = g [m₂ ( \frac{x}{d} -1) + m₁ ( \frac{L}{ 2d} -1) ]

c)  x = d - \frac{m_1}{m_2} \  \frac{L}{2d},  d)  m₂ = m₁  ( \frac{ L}{2d} -1)

Explanation:

After carefully reading your long sentence, I understand your exercise. In the attachment is a diagram of the assembly described. This is a balancing act

a) The tension of string A is requested

The expression for the rotational equilibrium taking the ends of the bar as the turning point, the counterclockwise rotations are positive

      ∑ τ = 0

      T_A d - W₂ x -W₁ L/2 = 0

      T_A = \frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )

b) the tension in string B

we write the expression of the translational equilibrium

       ∑ F = 0

       T_A - W₂ - W₁ - T_B = 0

       T_B = T_A -W₂ - W₁

       T_ B =   \frac{g}{d}\ ( m_2 x + m_1 \ \frac{L}{2} )  - g m₂ - g m₁

       T_B = g [m₂ ( \frac{x}{d} -1) + m₁ ( \frac{L}{ 2d} -1) ]

c) The minimum value of x for the system to remain stable, we use the expression for the endowment equilibrium, for this case the axis of rotation is the support point of the chord A, for which we will write the equation for this system

         T_A 0 + W₂ (d-x) - W₁ (L / 2-d) - T_B d = 0

at the point that begins to rotate T_B = 0

          g m₂ (d -x) -  g m₁  (0.5 L -d) + 0 = 0

          m₂ (d-x) = m₁ (0.5 L- d)

          m₂ x = m₂ d - m₁ (0.5 L- d)

          x = d - \frac{m_1}{m_2} \  \frac{L}{2d}

 

d) The mass of the block for which it is always in equilibrium

this is the mass for which x = 0

           0 = d - \frac{m_1}{m_2} \  \frac{L}{2d}

         \frac{m_1}{m_2} \ (0.5L -d) = d

          \frac{m_1}{m_2} = \frac{ d}{0.5L-d}

          m₂ = m₁  \frac{0.5 L -d}{d}

          m₂ = m₁  ( \frac{ L}{2d} -1)

5 0
2 years ago
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