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zhannawk [14.2K]
2 years ago
14

How long does it take electrons to get from

Physics
1 answer:
OlgaM077 [116]2 years ago
8 0

We need to find the time it takes an electron to move in the given circuit.

The time taken for electrons to reach the starting motor from the battery is 60.65 minutes.

I = Current = 134 A

N_A = Avogadro's number = 6.022\times 10^{23}\ \text{mol}^{-1}

A = Area = 38.9\ \text{mm}^2

L = Length = 92.2 cm

\rho = Density of copper = 8960\ \text{kg/m}^3

M = Molar mass of copper = 63.5 g/mol

n_v = Number of valence electrons of copper = 1

e = Charge of electron = 1.6\times 10^{-19}\ \text{C}

Number of charge carriers per unit volume is given by

n=\dfrac{\rho N_An_v}{M}\\\Rightarrow n=\dfrac{8960\times 6.022\times 10^{23}\times 1}{63.5\times 10^{-3}}\\\Rightarrow n=8.497\times 10^{28}\ \text{m}^{-3}

Time taken is given by

t=\dfrac{LAne}{I}\\\Rightarrow t=\dfrac{92.2\times 10^{-2}\times 38.9\times 10^{-6}\times 8.497\times 10^{28}\times 1.6\times 10^{-19}}{134}=3638.83\ \text{s}\\\Rightarrow t=\dfrac{3638.83}{60}=60.65\ \text{minutes}

The time taken for electrons to reach the starting motor from the battery is 60.65 minutes.

Learn more:

brainly.com/question/1426683

brainly.com/question/170663

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Which substance is translucent​?
liubo4ka [24]

Answer: C. Tissue Paper

Explanation:

When we talk about objects that are illuminated by a light source, they are classified according to the amount of light they let through them, as follows:

Transparent bodies: Those who let in almost all the light that incides them. Therefore, the intensity of the incident light is very similar to that transmitted.  For example: Clear Glass

Opaque bodies: Those who do not let the light pass.  For example: Cream Cheese and Orange Juice

Translucent bodies: Those that let in a portion of the incident light. That is, they let approximately half of the light that falls on them pass through. For example: Tissue Paper

5 0
4 years ago
A heliocentric system is _____-centered.<br><br> Milky Way<br> Earth<br> Moon<br> Sun
madreJ [45]

Answer: Sun

Explanation:

7 0
3 years ago
Read 2 more answers
An electron moves through a uniform electric field E = (2.00î + 5.40ĵ) V/m and a uniform magnetic field B = 0.400k T. Determine
EleoNora [17]

Answer:a=1.75\times 10^{21}\left ( 2\hat{i}5.08\hat{j}\right )

Explanation:

Given

Electric Field \vec{E}=2\hat{i}+5.4\hat{j}

\vec{B}=0.4\hat{k}\ T

velocity \vec{v}=8\hat{i} m/s

mass of electron m=9.1\times 10^{-31} kg

Force on a charge Particle moving in Magnetic Field

F=e\left [ \vec{E}+\left ( \vec{v}\times \vec{B}\right )\right ]

a=\frac{e\left [ \vec{E}+\left ( \vec{v}\times \vec{B}\right )\right ]}{m}

a=\frac{1.6\times 10^{-19}\left [ 2\hat{i}+5.4\hat{j}+\left ( 8\hat{i}\times 0.4\hat{k}\right )\right ]}{9.1\times 10^{-31}}  

a=1.75\times 10^{21}\left ( 2\hat{i}+5.08\hat{j}\right )\ m/s^2

7 0
3 years ago
1.A car starts from rest and acquires a velocity of 54 km/h in 2 minutes.Find(i) the acceleration and(ii) distance travelled by
Sergeu [11.5K]

Answer:

1.) 1620 km/h^2

2.) 2.7 km

Explanation:

1.) Given that the car start from rest. The initial velocity U will be equal to zero. That is,

U = 0.

Final velocity V = 54 km/h

Time t = 2 minute = 2/60 = 1/30 hour

Acceleration a will be change in velocity per time taken. That is,

a = ( V - U )/ t

Substitute V, U and t into the formula

a = 54 ÷ 1/30

a = 54 × 30 = 1620 km/h ^2

2.) Distance travelled S by the car during the time can be calculated by using the 2nd equation of motion.

S = Ut + 1/2at^2

Substitute all the parameters into the formula

S = 54 × 1/30 + 1/2 × 1620 × (1/30)^2

S = 54/30 + 810 × 1/900

S = 54/30 + 810/900

S = (1620+810)/900

S = 2430/900

S = 2.7 km.

Therefore, distance travelled by the car during this time is 2.7 km

4 0
3 years ago
Why do people use the scientific method?
Len [333]
When conducting research, scientists use the scientific method to collect measurable, empirical evidence in an experiment related to a hypothesis (often in the form of an if/then statement), the results aiming to support or contradict a theory.
Hopefully this helped.
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