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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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If a ball of mass, M, moving at velocity, v, collided with a ball of mass 10M at rest, describe what could happen to the velocit
KatRina [158]
Refer to the diagram shown below.

Before collision, the momentum of the two masses is
P₁ = Mv + (10M)*0 = Mv

After the collision, assume that the lighter ball rebounds off the heavier ball with a coefficient of restitution of r, so that v₂ = rv.
If r = 1, the rebound is elastic and v₂ = -v.
If r < 1, the rebound velocity is v₂ = -rv.
If r= 0, the lighter ball sticks to the heavier ball.

The momentum after collision is
P₂ = -Mv₂ + 10Mv₁

Because momentum is conserved, P₁ = P₂. That is,
10Mv₁ - M(rv) = Mv
v₁ = v(1+r)/10  for r>0.

When r=1 (elastic rebound)
v₁ = v/5.
The heavier ball moves right at 20% of the velocity of the lighter ball,
and the lighter ball rebounds with its velocity in the opposite direction.

When 0 < r < 1,
v₁ = (1+r)/10.
The heavier ball travels with greater than 20% of the velocity of the lighter ball, and the lighter ball rebounds with a velocity less than its initial velocity.

When r=0, the balls will stick together and
(10M + M)v₁ = Mv
v₁ = v/11.
The stuck balls move together at 1/11 of the initial velocity of the lighter ball.


3 0
3 years ago
In each cycle, a heat engine performs 710 J of work and exhausts 1480 J of heat. What is the thermal effiency?
Tomtit [17]

Answer:

32%

Explanation:

For a heat engine, efficiency is work out divided by heat in:

η = Wₒ / Qᵢ

Since energy is balanced, heat in is the sum of work out and heat out:

Qᵢ = Wₒ + Qₒ

Therefore:

η = Wₒ / (Wₒ + Qₒ)

Given Wₒ = 710 J and Qₒ = 1480 J:

η = 710 / (710 + 1480)

η = 0.32

The thermal efficiency is 32%.

6 0
3 years ago
Two small spheres, each carrying a net positive charge, are separated by 0.400 m. You have been asked to perform measurements th
neonofarm [45]

Answer:

a)  q₁ = 15. 28 10⁻⁶C, b)  q₂ = 5.64 10⁻⁶ C

Explanation:

For this exercise we use Newton's second law where force is Coulomb's electric force

Case 1. Distance (x₁ = 0.200 m) from the third sphere

         F₁ = F₁₃ - F₂₃

         F₁ = k q₁q₃ / x₁² - k q₂ q₃ / (0.4 - x₁)²

         F₁ = k q₃ (q₁ / x₁² - q₂ / (0.4- x₁)²

Case2 Distance (x₂ = 0.6 m) from the third sphere

        F₂ = F₁₃ + F₂₃

        F₂ = k q₁q₃ / x₂² + k q₂q₃ / (0.4- x₂)²

        F₂ = k q₃ (q₁ / x₂² + q₂ / (0.4-x₂)²

The distance is between the spheres, in the annex you can see the configuration of the charge and forces

Let's replace the values

        F₁ = 8.99 10⁹ 3.00 10⁻⁶⁶ (q₁ / 0.2² - q₂ / (0.4-0.2)²

        F₂ = 8.99 10⁹ 3.00 10⁻⁶ (q₁ / 0.6² + q₂ / (0.4-0.6)²

        6.50 = 674. 25 10³ (q₁ –q₂)

        3.50 = 26.97 10³ (q₁ / 0.36 + q₂ / 0.04)

We have a system of two equations with two unknowns, let's solve it. Let's clear q1 in the first and substitute in the second

         q₁ = q₂ + 6.50 / 674 10³

         3.50 / 26.97 10³ = (q₂ + 9.64 10⁻⁶) /0.36 + q₂ / 0.04

         1.2978 10⁻⁴ = q₂ / 0.36 + q₂ / 0.04 + 26.77 10⁻⁶

         q₂ (1 / 0.36 + 1 / 0.04) = 129.78 10⁻⁶ + 26.77 10⁻⁶

         q₂ 27,777 = 156,557 10⁻⁶

         q₂ = 156.557 10-6 /27.777

         q₂ = 5.636 10⁻⁶ C

We look for q1 in the other equation

        q₁ = q₂ + 6.50 / 674 10³

        q₁ = 5.636 10⁻⁶ + 9.6439 10⁻⁶

        q₁ = 15. 28 10⁻⁶C

4 0
3 years ago
A 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. what is its potential energy (PE) when it is 2.00 m above the grou
zloy xaker [14]

Answer:

1.5288 J

Explanation:

We can use the Energy conversion law

Energy can not be made or destroyed. It can only be converted to another form of energy

Potential Energy (PE) is the energy an object has due to its height of the location

<em>PE = mass * gravitational acceleration * height from the ground</em>

Simply here , lemming had a PE and KE also.  

(g = 9.8 m/s²)

PE = 0.0780 * 9.8  *2

     = 1.5288 J

4 0
3 years ago
Explain the following observation referring to the appropriate process of heat transfer in each case
tigry1 [53]
If the two cups are the same/equal in all regards the spoon makes the difference in heat loss. Metal spoon is a good conductor, heat conducts up the spoon handle. Radiant and convection losses occur in the surrounding air. Metal becomes a “heat sink”.

Spoon will feel warmer to the touch than a pottery or foam cup that is not a good conductor.
5 0
3 years ago
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