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KonstantinChe [14]
3 years ago
15

Need help use screenshot(SAT Prep) In △ABC, AB = BC = 20 and DE ≈ 9.28. Approximate BD. bd equals?

Mathematics
1 answer:
Lilit [14]3 years ago
4 0

Answer:

5.36

Step-by-step explanation:

Given that:

<BAD = <CAE, therefore, BD = EC

Let's take x to be the length of BD = EC

BD + DE + EC = BC

BC = 20,

BD = EC = x

DE ≈ 9.28

Thus,

x + 9.28 + x = 20

x + x + 9.28 = 20

2x + 9.28 = 20

Subtract 9.28 from both sides

2x + 9.28 - 9.28 = 20 - 9.28

2x = 10.72

Divided both sides by 2 to solve for x

\frac{2x}{2} = \frac{10.72}{2}

x = \frac{10.72}{2}

x = 5.36

BD ≈ 5.36

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conservation of angular momentum ; conservation of energy

Step-by-step explanation:

The complete Question is given as follows:

" A uniform bar of mass (m) and length (L) is suspended on a frictionless hinge. A horizontally launched blob of clay of mass (m) strikes the bottom end of the bar and sticks to it. After that, the bar swings upward. What is the minimum initial speed (v) of the blob of clay that would enable the rod to swing a full circle? Which concepts/laws would be most helpful in solving this problem? Select the best answer from the options below.  "

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kinematics of rotational motion; conservation of energy

conservation of momentum ; conservation of energy

conservation of angular momentum ; conservation of momentum

conservation of angular momentum ; conservation of energy

conservation of energy ; Newton's laws

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Solution:

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                                       M_i = M_f

- Initially the rod was at rest and clay had velocity of v, then M_i can be written as:

                                       M_i = m*v*L

- The final momentum is the combined effect of clay and rod:

                                       M_f = ( m*L^2 + I_rod )*w

- Where w is the angular speed of the rod after impact. And I_rod is the moment of inertia of rod.

                                      I_rod = mL^2 / 3

                                      M_f = ( m*L^2 + m*L^2 /3 )*w = (4*m*L^2 / 3)*w

- Formulate w in terms of initial velocity v:

                                     m*v*L = (4*m*L^2 / 3)*w

                                      0.75*v / L = w

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- Apply conservation of Energy principle:

                                     T_i + V_i = T_f + V_f

Where, T is the kinetic energy soon after impact and at top most position.

            Assuming, T_f = 0 , for minimum velocity required to complete on circle.

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             T_i = 0.5*[m]*[v^2]

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