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I am Lyosha [343]
3 years ago
12

You are given a copper bar of dimensions 3 cm × 5 cm × 8 cm and asked to attach leads to it in order to make a resistor. (a) If

you want to achieve the SMALLEST possible resistance, you should attach the leads to the opposite faces that measure A) 3 cm × 5 cm. B) 3 cm × 8 cm. C) 5 cm × 8 cm. D) Any pair of faces produces the same resistance. (b) If you want to achieve the LARGEST possible resistance, you should attach the leads to the opposite faces that measure A) 3 cm × 5 cm. B) 3 cm × 8 cm. C) 5 cm × 8 cm. D) Any pair of faces produces the same resistance.
Physics
1 answer:
dem82 [27]3 years ago
4 0

Answer:

A) If you want to achieve the SMALLEST possible resistance, you should attach the leads to the opposite faces that measure b) 5 cm by 8 cm.

B) If you want to achieve the LARGEST possible resistance, you should attach the leads to the opposite faces that measure a) 3 cm × 5 cm

Explanation:

Resistivity is directly proportional to lenght and inversely properly to cross sectional area.

For the first case, 5 cm by 8 cm gives the largest area and leave 3 cm as the lenght. The resistivity of the metal will be smallest in these dimensions.

For the second case, 3 cm by 5 cm gives the smallest area, leaving 8 cm as the lenght. This is the maximum arrangement that can give the largest resistance possible.

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Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg),
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Answer:

v=3.564\ m.s^{-1}

\Delta v =2.16\ m.s^{-1}

Explanation:

Given:

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height between John and William, h=1.8\ m

<u>Using the equation of motion:</u>

v_J^2=u_J^2+2 (g.sin\theta).l

where:

v_J = final velocity of John at the end of the slide

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Now putting respective :

v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3

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<u>Now using the law of conservation of momentum at the bottom of the slide:</u>

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m_J.v_J+m_w.v_w=(m_J+m_w).v

where: v = velocity with which they move together after collision

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(B)

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<u>Now we find the force along the slide due to the body weight:</u>

F=m_J.g.sin\theta

F=30\times 9.8\times \frac{1.8}{3}

F=176.4\ N

<em><u>Hence the net force along the slide:</u></em>

F_R=71.4\ N

<em>Now the acceleration of John:</em>

a_j=\frac{F_R}{m_J}

a_j=\frac{71.4}{30}

a_j=2.38\ m.s^{-2}

<u>Now the new velocity:</u>

v_J_n^2=u_J^2+2.(a_j).l

v_J_n^2=0^2+2\times 2.38\times 3

v_J_n=3.78\ m.s^{-1}

Hence the new velocity is slower by

\Delta v =(v_J-v_J_n)

\Delta v =5.94-3.78= 2.16\ m.s^{-1}

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