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vagabundo [1.1K]
2 years ago
3

High-voltage power lines are a familiar sight throughout the country. The aluminum wire used for some of these lines has a cross

-sectional area of 5.00 10-4 m2. What is the resistance of ten kilometers of this wire?
Physics
1 answer:
ale4655 [162]2 years ago
6 0

Answer:

Resistance, R = 0.564 ohms

Explanation:

Area of cross section of the aluminium wire, A=5\times 10^{-4}\ m^2

Length of the wire, L = 10 km

Resistivity of the aluminium wire is, \rho=2.82\times 10^{-8}\ \Omega-m

The resistivity of a wire is given by the relation as follows :

R=\rho\dfrac{L}{A}\\\\R=2.82\times 10^{-8}\times \dfrac{10\times 10^3}{5\times 10^{-4}}\\\\R=0.564\ \Omega

So, the resistance of the wire is 0.564 ohms.        

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Which type of listening response includes the use of head nods, facial expressions, and short utterances such as "uh-huh" that s
BARSIC [14]
This type of listening response is called back-channel signal. This allows the speaker to know that the listener is attentive or willing to engage a conversation between them. It is shown through short utterances, facial expressions, head nods and others. 
4 0
2 years ago
Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea
madreJ [45]
In order to answer this exercise you need to use the formulas

 S = Vo*t + (1/2)*a*t^2

Vf = Vo + at

The data will be given as

Vf = final velocity = ?

Vo = initial velocity = 1.4 m/s

a = acceleration = 0.20 m/s^2

s = displacement = 100m

And now you do the following:

100 = 1.4t + (1/2)*0.2*t^2

t = 25.388s

and

Vf = 1.4 + 0.2(25.388)

Vf = 6.5 m/s

So the answer you are looking for is 6.5 m/s
7 0
2 years ago
Read 2 more answers
You testify as an expert witness in a case involving an accident in which car A slid into the rear of car B, which was stopped a
bekas [8.4K]

Answer:

A) 12.08 m/s

B) 19.39 m/s

Explanation:

A) Down the hill, we will apply Newton’s second law of motion in the downward direction to get:

mg(sinθ) – F_k = ma

Where; F_k is frictional force due to kinetic friction given by the formula;

F_k = (μ_k) × F_n

F_n is normal force given by mgcosθ

Thus;

F_k = μ_k(mg cosθ)

We now have;

mg(sinθ) – μ_k(mg cosθ) = ma

Dividing through by m to get;

g(sinθ) – μ_k(g cosθ) = a

a = 9.8(sin 12.03) - 0.6(9.8 × cos 12.03)

a = -3.71 m/s²

We are told that distance d = 24.0 m and v_o = 18 m/s

Using newton's 3rd equation of motion, we have;

v = √(v_o² + 2ad)

v = √(18² + (2 × -3.71 × 24))

v = 12.08 m/s

B) Now, μ_k = 0.10

Thus;

a = 9.8(sin 12.03) - 0.1(9.8 × cos 12.03)

a = 1.08 m/s²

Using newton's 3rd equation of motion, we have;

v = √(v_o + 2ad)

v = √(18² + (2 × 1.08 × 24))

v = 19.39 m/s

6 0
2 years ago
A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.
fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

F=\dfrac{mv^2}{r}-mg

F=m(\dfrac{v^2}{r}-g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)

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(b) The net tension in the string when the ball is at the bottom of the circle is given by :

F=\dfrac{mv^2}{r}+mg

F=m(\dfrac{v^2}{r}+g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}

T=\dfrac{m}{r}(g(r-h)+v^2)

Since, v^2=u^2-2gh

T=\dfrac{m}{r}(u^2-3gh+gr)

Hence, this is the required solution.

6 0
2 years ago
The structural diversity of carbon-based molecules is determined by which properties?
Leokris [45]

Explanation:

The structural diversity of carbon-based molecules is determined by following properties:

1. the ability of those bonds to rotate freely,

2.the ability of carbon to form four covalent bonds,

3.the orientation of those bonds in the form of a tetrahedron.

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