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Korvikt [17]
3 years ago
8

Birds resting on high-voltage power lines are a common sight. the copper wire on which a bird stands is 1.28 cm in diameter and

carries a current of 149
a. if the bird's feet are 4.12 cm apart, what is the potential difference across its body? copper's resistivity is 1.68 × 10−8 ω m . answer in units of µv.
Physics
1 answer:
nekit [7.7K]3 years ago
4 0

Answer:

8\cdot 10^{-4} V

Explanation:

First of all, let's find the cross-sectional area of the copper wire. The radius of the wire half the diameter:

r=\frac{d}{2}=\frac{1.28 cm}{2}=0.64 cm=6.4\cdot 10^{-3} m

So the area is

A=\pi r^2 = \pi (6.4\cdot 10^{-3} m)^2=1.29\cdot 10^{-4} m^2

Now we can calculate the resistance of the piece of copper wire between the bird's feet, with the formula:

R=\rho \frac{L}{A}

where

\rho=1.68\cdot 10^{-8} \Omega m is the resistivity of copper

L=4.12 cm=4.12 \cdot 10^{-2} m is the length of the piece of wire

A=1.29\cdot 10^{-4} m^2 is the cross-sectional area

Substituting, we find

R=(1.68\cdot 10^{-8} m^2)\frac{4.12\cdot 10^{-2} m}{1.29\cdot 10^{-4} m^2}=5.4\cdot 10^{-6} \Omega

And since we know the current in the wire, I=149 A, we can now find the potential difference across the body of the bird, by using Ohm's law:

V=IR=(149 A)(5.4\cdot 10^{-6} \Omega)=8\cdot 10^{-4} V

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According to Hubble's law

v=H_0D\\\Rightarrow D=\frac{v}{H_0}\\\Rightarrow D=\frac{3\times 10^{4}}{65}\times 3.2\times 10^6\\\Rightarrow D=1476923076.92307\ ly\\\Rightarrow D=1.4\times 10^9\ ly

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5 0
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You are travelling by skateboard at 3.0 m/s and start to accelerate. If you
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Answer: 8m/s

Explanation:

Vs= 3 m/s

Vf=?

a=0.5m/s²

t=10s

-----------

a=Vf-Vs/t

at=Vf-Vs

0.5*10s=Vf-Vs

5m/s=Vf-3m/s

5m/s+3m/s=Vf

Vf=8m/s

6 0
3 years ago
A glass rod is rubbed briefly with silk and the rod becomes able to attract and lift a number of small pieces of paper. If the r
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9. The Metro Rail makes a stop at NRG Stadium to pick up some Texans fans after the game. After
attashe74 [19]

Answer:

It takes the train 7.5 seconds to reach top speed

The train travels 112.5 meters before it reaches its top speed

Explanation:

The Metro Rail makes a stop at NRG Stadium to pick up some Texans

fans after the game

After  all the passengers board, the train accelerates at a rate of 4 m/s²

to a top speed of 30 m/s

We need to find the train takes how many seconds to reach the

top speed

Given is:

→ The train start from rest, then its initial speed u = 0

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→ Its top speed v = 30 m/s

To find the time we can use this rule;

→ t=\frac{v-u}{a}

→ t=\frac{30-0}{4}=\frac{30}{4}=7.5 seconds

<em>It takes the train 7.5 seconds to reach top speed</em>

<em></em>

Now we need to find how far the train travels before reaches top speed

We can use this rule;

→ v² = u² + 2as, where s is the distance

→ v = 30 m/s , u = 0 , a = 4 m/s²

Substitute these values in the rule

→ (30)² = 0 + 2(4) s

→ 900 = 8 s

Divide both sides by 8

→ s = 112.5 m

<em>The train travels 112.5 meters before it reaches its top speed </em>

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Answer:

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