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zhuklara [117]
3 years ago
12

The resistance, R, of a wire varies directly as its length and inversely as the square of its diameter. If the resistance of a w

ire 2700ft long with a diameter of 0.26 inches is 9850 ohms, what is the resistance of 2800ft of the same type of wire with a diameter of 0.1 inches? (Leave k in fraction form or round to at least 3 decimal places. Round off your final answer to the nearest hundredth.)
Physics
1 answer:
lbvjy [14]3 years ago
8 0

R is proportional to the length of the wire:

R ∝ length

R is also proportional to the inverse square of the diameter:

R ∝ 1/diameter²

The resistance of a wire 2700ft long with a diameter of 0.26in is 9850Ω. Now let's change the shape of the wire, adding and subtracting material as we go along, such that the wire is now 2800ft and has a diameter of 0.1in.

Calculate the scale factor due to the changed length:

k₁ = 2800/2700 = 1.037

Scale factor due to changed diameter:

k₂ = 1/(0.1/0.26)² = 6.76

Multiply the original resistance by these factors to get the new resistance:

R = R₀k₁k₂

R₀ = 9850Ω, k₁ = 1.037, k₂ = 6.76

R = 9850(1.037)(6.76)

R = 69049.682Ω

Round to the nearest hundredth:

R = 69049.68Ω

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GalinKa [24]

Answer:

a)  t = 0.90 s, b)  t = 0.815 s, c)  t = 0.90 s, d)  x = 3.6 m, e)  t = 0.639 s

Explanation:

all these exercises are about kinematics

a) The body is released from rest,  

           y = y₀ + v₀ t - ½ g t²

in this case when reaching the ground y = 0 and its initial velocity is vo = 0

           0 = y₀ + 0 - ½ g t²

           t² = 2 y₀ / g

           t² = 2 4 /9.81

           t² = 0.815

            t = √0.815

           t = 0.90 s

b) It is thrown upwards at v₀ = 4 m / s

         y = y₀ + v₀ t - ½ g t²

in this case the initial and final height is the same

        y = y₀ = 0

        0 = v₀ t -1/2 g t²

        t = 2 v₀ / g

        t = 2 4 /9.81

        t = 0.815 s

c) the ball is at y₀ = 4 m and its initial velocity is horizontal v₀ = 4 m / s

        y = y₀ + v_{oy} t - ½ g t²

        0 = y₀ + 0 - ½ g t²

        t² = 2 i / g

        t² = 2 4 / 9.81

        t² = 0.815

        t = 0.90 s

d) the horizontal distance traveled is

        x = v₀ₓ t

        x = 4 0.90

        x = 3.6 m

e) We can calculate the time to fall from I = 2 m

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        0 = y₀ + 0 - ½ g t²

        t² = 2 y₀i / g

        t² = 2 2 /9.81

        t² = 0.4077

        t = 0.639 s

Therefore, when making measurements, you should find readings around this value.

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Answer:

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