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777dan777 [17]
3 years ago
14

A wire is 1.1 m long and 0.91 mm in cross-sectional area. It carries a current of 4.4 A when a 1.8 V potential difference is app

lied between its ends. Calculate the conductivity of the material of which this wire is made. Number Units The number of significant digits is set to 2; the tolerance is +/-2%
Physics
1 answer:
Vlad [161]3 years ago
3 0

Answer:

The conductivity of the material from which the conductor is made is 2.95 x10⁶  [S. m⁻¹]  .

Explanation:

L= 1.1m

S= 0.91 mm²= 9.1 x10⁻⁷ m²

I= 4.4 A

V= 1.8V

\frac{V}{I} =R\\\\\frac{1.8V}{4.4A} =R= 9/22 ohms

R=9/22Ω

R=\frac{L}{σ . S} \\\\Conductivity=sigma=\frac{L}{R.S} \\\\Conductivity= \frac{1.1m}{9/22 ohms . 9.1 E10 (-7) square meters}

Conductivity=2.95 x10⁶  [S. m⁻¹]

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You jump off a truck and accelerate toward the surface of the Earth. Does the Earth accelerate toward you?
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To find the answer, we need to know about the acceleration of earth due to the gravitational attraction.

<h3>What's the gravitational force between the earth and a person?</h3>
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7 0
2 years ago
A loop of area 0.250 m^2 is in a uniform 0.020 0-T magnetic field. If the flux through the loop is 3.83 × 10-3T· m2, what angle
dangina [55]

Answer:

40.0⁰

Explanation:

The formula for calculating the magnetic flux is expressed as:

\phi = BAcos\theta where:

\phi is the magnetic flux

B is the magnetic field

A is the cross sectional area

\theta is the angle that the normal to the plane of the loop make with the direction of the magnetic field.

Given

A = 0.250m²

B = 0.020T

\phi = 3.83 × 10⁻³T· m²

3.83 × 10⁻³ = 0.020*0.250cosθ

3.83 × 10⁻³ = 0.005cosθ

cosθ = 0.00383/0.005

cosθ = 0.766

θ = cos⁻¹0.766

θ = 40.0⁰

<em>Hence the angle normal to the plane of the loop make with the direction of the magnetic field is 40.0⁰</em>

4 0
3 years ago
Two parallel-plate capacitors have the same plate area, but the plate gap in capacitor 1 is twice as big as capacitor 2. If capa
-BARSIC- [3]

Answer:

Capacitance of the second capacitor = 2C

Explanation:

\texttt{Capacitance, C}=\frac{\varepsilon_0A}{d}

Where A is the area, d is the gap between plates and ε₀ is the dielectric constant.

Let C₁ be the capacitance of first capacitor with area A₁ and gap between plates d₁.

We have    

              \texttt{Capacitance, C}_1=\frac{\varepsilon_0A_1}{d_1}=C

Similarly for capacitor 2

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Capacitance of the second capacitor = 2C

6 0
3 years ago
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