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NemiM [27]
3 years ago
6

A metallic conductor has a resistivity of 35 × 10 −6 Ω⋅m. What is the resistance of a piece that is 20 m long and has a uniform

cross-sectional area of 4.0 × 10 −6 m 2?
Physics
1 answer:
Nitella [24]3 years ago
3 0

Answer:

Therefore the resistance of the conductor is 175Ω

Explanation:

Resistance:

  • Resistance of a metallic conductor is directly proportional to its length(l).
  • Resistance of a metallic conductor is inversely proportional to its cross section area(A).

The notation sign of resistance is R.

The unit of resistance is ohm (Ω).

Therefore,

R \propto l

and

R \propto \frac{1}{A}

\therefore R\propto \frac{l}{A}

\Rightarrow R=\rho \frac{l}{A}

ρ is the proportional constant.

It is also known as resistivity of that metal.

Given ρ=35×10⁻⁶Ω-m

l= 20 m

A= 4.0×10⁻⁶m²

\therefore R=35\times 10^{-6}\times \frac{20}{4.0\times 10^{-6}}

       =175Ω

Therefore the resistance of the conductor is 175Ω

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In this activity, you will research how scientists developed historical
Darya [45]

Answer: Bohr's model (1913)

  Niels Bohr improved Rutherford's model. Using mathematical ideas, he showed that electrons occupy shells or energy levels around the nucleus. The Dalton model has changed over time because of the discovery of subatomic particles.

Bohr's model (1913)

  Niels Bohr improved Rutherford's model. Using mathematical ideas, he showed that electrons occupy shells or energy levels around the nucleus. The Dalton model has changed over time because of the discovery of subatomic particles

  Dalton's atomic theory proposed that all matter was composed of atoms, indivisible and indestructible building blocks. While all atoms of an element were identical, different elements had atoms of differing size and mass.

  John Dalton

The idea that everything is made of atoms was pioneered by John Dalton (1766-1844) in a book he published in 1808. He is sometimes called the "father" of atomic theory, but judging from this photo on the right "grandfather" might be a better term.

Explanation:

8 0
3 years ago
Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ
Irina-Kira [14]

Answer:

Magnitude of the net force on q₁-

Fn₁=1403 N

Magnitude of the net force on q₂+

Fn₂= 810 N

Magnitude of the net force on q₃+

Fn₃= 810 N

Explanation:

Look at the attached graphic:

The charges of the same sign exert forces of repulsion and the charges of   opposite sign exert forces of attraction.

Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:

F= (k*q*q)/(d)²

F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N

Magnitude of the net force on q₁-

Fn₁x= 0

Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N

Fn₁=1403 N

Magnitude of the net force on q₃+

Fn₃x= 810- 810 cos 60° = 405 N

Fn₃y= 810*sin 60° = 701.5 N

Fn_{3} = \sqrt{405^{2}+701.5^{2}  }

Fn₃ = 810 N

Magnitude of the net force on q₂+

Fn₂ = Fn₃ = 810 N

6 0
3 years ago
In traveling a distance of 2.5 km between points A and D, a car is driven at 99 km/h from A to B for t seconds and 48 km/h from
Andreyy89

Answer:

d = 1.954 Km

Explanation:

given,

total distance, D = 2.5 Km

in stretch A to B =

speed = 99 Km/h = 99 x 0.278 = 27.22 m/s     time =t

in stretch B to C

time = 3.4 s

In stretch C to D

speed = 48 Km/h = 48 x 0.278 = 13.34 m/s     time =t      

we know,

distance = speed x time

distance of BC

using equation of motion

v = u + a t

27.22 = 13.34 - a x 3.4

a = 4.08 m/s²

uniform deceleration is equal to 4.08 m/s²

distance traveled in BC

s = ut + \dfrac{1}{2}at^2

s = 13.34\times 3.4 + \dfrac{1}{2}\times 4.08 \times 3.4^2

s = 68.94 m

3000 = 99 \times \dfrac{1000\ t}{3600}+ 68.94 + 48\times \dfrac{1000\ t}{3600}

3000 = 27.5 t + 68.94 + 13.33 t

40.83 t = 2931.06

t = 71.79 s

distance travel in AB

distance = s x t

d = 27.22 x 71.79

d = 1954 m

d = 1.954 Km

distance between A and B is equal to 1.954 Km.

4 0
3 years ago
Why are most substances denser in the solid state than in the liquid state?
Rom4ik [11]

Answer:

For other liquids, solidification when the temperature drops includes the lowering of kinetic energy, which allows molecules to pack more tightly and makes the solid denser than its liquid form. Because ice is less dense than water, it is able to float at the surface of water.

Explanation:

have a great day:)

3 0
2 years ago
Three charges 1.5*10-6, 3*10-6, -3*10-6 are placed at three vertices of an equilateral triangle of side 30cm. Find the net force
gulaghasi [49]

Answer:

F = 0N

Explanation:

The force between two charges is given by

F=k\frac{q_1q_2}{r^2}

where r is the distance between the charges and K is the Coulomb's constant

(k=8-89*10^9Nm^2/C^2)

The force in the first charge is only the sum of the forces due to the other charges. Hence we have

F_T=F_1+F_2=k\frac{q_2q_1}{r^2}+k\frac{q_3q_1}{r^2}

F_T=(8.89*10^9\frac{Nm^2}{C^2})\frac{(3*10^{-6}C)(1.5*10^{-6}C)}{(0.3m)^2}+(8.89*10^9\frac{Nm^2}{C^2})\frac{(-3*10^{-6}C)(1.5*10^{-6}C)}{(0.3m)^2}\\\\F_T=0.445N-0.445N=0N

Ft=0N

Hope this helps!!

5 0
3 years ago
Read 2 more answers
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