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Andrew [12]
3 years ago
14

The length of an aluminum wire is quadrupled and the radius is doubled. By which factor does the resistance change?

Physics
1 answer:
erik [133]3 years ago
7 0
The resistance of a conductive wire is given by:
R= \frac{\rho L}{A}
where
\rho is the material resistivity
L is the wire length
A is the cross-sectional area of the wire

The length of the wire is quadrupled, so if we call L the original length and L' the new length, we can write 
L'=4 L

Similarly, the radius of the wire is doubled (r'=2r), so the new area is
A'= \pi (r')^2 = \pi (2r)^2 = 4 \pi r^2 = 4A

And if we substitute into the equation, we find that the new resistance of the wire is
R'= \frac{\rho L'}{A'}= \frac{\rho (4L)}{4 A'}  =  \frac{\rho L}{A}=R
Therefore, R=R': this means that the resistance of the wire did not change.
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The distance covered by a car moving with a speed of 36 km/h in 15m is (a) 0.9 km (b) 9.0 km (C) 90 km (d) 900 km​
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This is a uniform rectilinear motion exercise.

To start solving this exercise, we obtain the following data:

<h3>Data:</h3>
  • v = 36 km/h
  • d = ¿?
  • t = 15 min = 0.25 hr

                                     Minutes to hour converter

                                                  60 min = 1 hour

                                       \boldsymbol{\sf15\not{min}*\dfrac{1 \ hr}{60 \not{min}}=0.25 \ hr  }}

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We apply the following formula: d = v * t

We substitute our data into the formula:

                                     \boldsymbol{\sf{d=v*t}}

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(a) The velocity (v)  of the ball when reaches its maximum altitude is zero .

     v= 0

(b) The acceleration of an object free fall motion is constant and is equal to the acceleration due to gravity,then, At maximum height the acceleration of ball  is  g =-9,8 m/s².

(c)velocity of the ball when it returns to the ground :

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Explanation:

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vf²=v₀²+2*a*y Formula (1)

y:displacement in meters (m)  

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration

Data

v₀ = +5.00 m/s

Problem development

(a) What is its velocity when it reaches its maximum altitude?

in ymax,  vf=0 , At maximum height the velocity is zero and the ball falls freely

(b) What is its acceleration at this point?

The acceleration of an object free fall motion is constant and is equal to the acceleration due to gravity,then, At maximum height the acceleration of ball  is 9,8 m/s².

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a is constant , a= g = -9.8 m/s²

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