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uranmaximum [27]
3 years ago
13

How long does it take electrons to get from the car battery to the starting motor? Assume the current is 143 A and the electrons

travel through copper wire with cross sectional area 40.2 mm2 and length 84.5 cm . The mass density of copper is 8960 kg/m3 and the molar mass is 63.5 g/mol . Avogadro’s number is 6.022 × 1023 . Assume that each copper atom contributes one conduction electron.
Physics
1 answer:
Ostrovityanka [42]3 years ago
5 0

Answer

the concentration of electron/ volume (n) = 8.5 × 10²⁸ /m³

the drift velocity of electron

v = \dfrac{I}{neA}

v = \dfrac{143}{8.5\times 10^{28}\times 1.6\times 10^{-19}\times 40.2 \times 10^{-6}}

v = 2.615 × 10⁻⁴ m/s

time require is

t=\dfrac{L}{v}

t=\dfrac{0.845}{2.615\times 10^{-4}}

t = 3231.35 s

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Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at
PSYCHO15rus [73]

Answer:

|\Delta v |=\sqrt{\frac{4Gm}{d} }

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

\mu= \frac{m_1m_2}{m_1+m_2}

now, since both the masses have mass m

therefore,

\mu= \frac{m^2}{2m}

= m/2

The final K.E. of the particles is

KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2

Distance between two particles is d and the gravitational potential energy between them is given by

PE_{Gravitational}= \frac{Gmm}{d}

By law of conservation of energy we have

KE_{initial}+KE_{final}= PE_{gravitaional}

Now plugging the values we get

0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}

|\Delta v |=\sqrt{\frac{4Gm}{d} }

=\sqrt{\frac{Gm}{d} }

This the required relation between G,m and d

5 0
2 years ago
What are the answers please help
aleksandr82 [10.1K]

Answer:

a.) The main scale reading is 10.2cm

b.) Division 7 = 0.07

c.) 10.27 cm

d.)  10.31 cm

e.)  10.24 cm  

Explanation:

The figure depicts a vernier caliper readings

a.) The main scale reading is 10.2 cm

The reading before the vernier scale

b.) Division 7 = 0.07

the point where the main scale and vernier scale meet

c.) The observed readings is

10.2 + 0.07 = 10.27 cm

d.) If the instrument has a positive zero error of 4 division

correct reading = 10.27 + 0.04 = 10.31cm

e.)  If the instrument has a negative zero error of 3 division

correct reading = 10.27 - 0.03 = 10.24cm  

3 0
3 years ago
You wrap a wire around a piece of iron. If you slowly increase the strength of an electric current flowing through the wire, you
Maksim231197 [3]
The correct answer is A. the magnet to become stronger

The stronger the electric current in the piece of metal, the stronger the magnetic field will be.
4 0
3 years ago
Read 2 more answers
You are pulling a child in a wagon. The rope handle is inclined upward at a 60∘ angle. The tension in the handle is 20 N.
dem82 [27]
  • Angle (θ) = 60°
  • Force (F) = 20 N
  • Distance (s) = 200 m
  • Therefore, work done
  • = Fs Cos θ
  • = (20 × 200 × Cos 60°) J
  • = (20 × 200 × 1/2) J
  • = (20 × 100) J
  • = 2000 J

<u>Answer</u><u>:</u>

<u>2</u><u>0</u><u>0</u><u>0</u><u> </u><u>J</u>

Hope you could get an idea from here.

Doubt clarification - use comment section.

6 0
2 years ago
Read 2 more answers
Inside a television picture tube there is a build-up of electrons (charge of 1.602 × 10^–19 C) with an average spacing of 38.0 n
Brut [27]
The magnitude of electric field is produced by the electrons at a certain distance.

E = kQ/r²

where: 
E = electric field produced
Q = charge
r = distance
k = Coulomb Law constant 9 x10^9<span> N. m</span>2<span> / C</span><span>2

Given are the following:
Q = </span><span>1.602 × 10^–19 C
</span><span>r = 38 x 10^-9 m

Substitue the given:
E = </span>\frac{( 1.602 x 10^{-19} )( 9.0x10_{9} )}{(38x10^{-9}) ^{2} }

E = 998.476 kN/C


8 0
3 years ago
Read 2 more answers
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