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KengaRu [80]
3 years ago
12

A 230-km-long high-voltage transmission line 2.0 cm in diameter carries a steady current of 1,100 A. If the conductor is copper

with a free charge density of 8.5 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable? (Use 3.156 107 for the number of seconds in a year.)
Physics
2 answers:
Nadusha1986 [10]3 years ago
7 0

Answer:

28.23 years

Explanation:

It takes 28.23 years to take one electron to travel the full length of the cable.

3.156 107 = number of seconds in a year.

230000 / (2.58 x 10^-4) = 8.91 x 10^8 second.

Molodets [167]3 years ago
5 0

Answer:

28.23 years

Explanation:

I = 1100 A

L = 230 km = 230, 000 m

diameter = 2 cm

radius, r = 1 cm = 0.01 m

Area, A = 3.14 x 0.01 x 0.01 = 3.14 x 10^-4 m^2

n = 8.5 x 10^28 per cubic metre

Use the relation

I = n e A vd

vd = I / n e A

vd = 1100 / (8.5 x 10^28 x 1.6 x 10^-19 x 3.14 x 10^-4)

vd = 2.58 x 10^-4 m/s

Let time taken is t.

Distance = velocity x time

t = distance / velocity = L / vd

t = 230000 / (2.58 x 10^-4) = 8.91 x 10^8 second

t = 28.23 years

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Answer

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d=\frac{\Delta V}{E}=\frac{1.5 V}{100 V/m}=0.015 m=1.5 \cdot 10^{-2} m

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A mechanic needs to replace the motor for a merry-go-round. The merry-go-round should accelerate from rest to 1.5 rad/s in 6.0s
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109656.25 Nm

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\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\dfrac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\dfrac{1.5-0}{6}\\\Rightarrow \alpha=0.25\ rad/s^2

Torque is given by

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The torque specifications must be 109656.25 Nm

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Nuclearissionoccursthroughmanydifferent pathways. For the ission of U-235 induced by a neutron, write a nuclear equation to form
ruslelena [56]

Answer: a) ^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n

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Explanation:

A nuclear fission reaction is defined as the reaction in which a heavy nucleus splits into small nuclei along with release of energy.

a) The given reaction is ^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + x^{1}_{0}n

Now,  as the mass on both reactant and product side must be equal:

235+1=87+146+x

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Thus three neutrons are produced and nuclear equation will be: ^{235}_{92}\textrm{U}+^{1}_{0}\textrm{n} \rightarrow ^{87}_{35}\textrm {Br}+ ^{146}_{57}\La + 3^{1}_{0}n

b) For the another fission reaction:

^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{A}_{56}\textrm{Ba}+^{94}_{Z}\textrm{X}+2^1_0\textrm{n}

To calculate A:

Total mass on reactant side = total mass on product side

235 + 1 = A + 94 + 2

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To calculate Z:

Total atomic number on reactant side = total atomic number on product side

92 + 0 = 56 + Z + 0

Z = 36

As Krypton has atomic number of 36,Thus the nuclear equation will be :

^{235}_{92}\textrm{U}+^1_0\textrm{n}\rightarrow ^{140}_{56}\textrm{Ba}+^{94}_{36}\textrm{Kr}+2^1_0\textrm{n}

         

8 0
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