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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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Answer:

The Answer is A They can damage hearing.

Explanation:

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3 years ago
A quantity that has direction as well as size is a scalar.<br><br><br> True or False
lisov135 [29]

Answer:

A quantity that does not depend on the direction is called a scalar quantity. Vector quantities have two characteristics, a magnitude, and a direction. Scalar quantities have only a magnitude. When comparing two vector quantities of the same type, you have to compare both the magnitude and the direction.

Scalar quantities only have magnitude (size). Scalar quantities include distance...

A quantity that is specified by both size and direction is a vector. Displacement includes both size and direction and is an example of a vector. However, distance is a physical quantity that does not include a direction and isn't a vector.

Explanation:

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3 years ago
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3.) An engineer is designing the runway for an airport. Of the planes that will use the airport,
scoray [572]

Answer: 704

Explanation:Vi = 0 m/s

vf = 65 m/s

a = 3 m/s2

d = ??

vf2 = vi2 + 2*a*d

(65 m/s)2 = (0 m/s)2 + 2*(3 m/s2)*d

4225 m2/s2 = (0 m/s)2 + (6 m/s2)*d

(4225 m 2/m2)/(6 m/s2) = d

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3 years ago
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g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the
MrRa [10]

Answer:

Approximately -1.58\; \rm rad \cdot s^{-2}.

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

v^2 - u^2 = 2\, a\, x,

where

  • v is the final velocity of the moving object,
  • u is the initial velocity of the moving object,
  • a is the (linear) acceleration of the moving object, and
  • x is the (linear) displacement of the object while its velocity changed from u to v.

The angular analogue of that equation will be:

(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta, where

  • \omega(\text{final}) and \omega(\text{initial}) are the initial and final angular velocity of the rotating object,
  • \alpha is the angular acceleration of the moving object, and
  • \theta is the angular displacement of the object while its angular velocity changed from \omega(\text{initial}) to \omega(\text{final}).

For this object:

  • \omega(\text{final}) = 0\; \rm rad\cdot s^{-1}, whereas
  • \omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}.

The question is asking for an angular acceleration with the unit \rm rad \cdot s^{-1}. However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}.

Rearrange the equation (\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta and solve for \alpha:

\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}.

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Answer:pelo o que eu sei é ..

V

V

V

F

F

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Explanation:

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