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kobusy [5.1K]
3 years ago
15

If the vertical posts supporting these wires divide the lines into straight 19.0 m segments, what magnetic force does each segme

nt exert on the other?

Physics
1 answer:
erica [24]3 years ago
5 0

Answer:

Complete question is:

Two high-current transmission lines carry currents of 29.0 A and 78.0 A in the same direction and are suspended parallel to each other 38.0 cm apart.

Part A.) If the vertical posts supporting these wires divide the lines into straight 20.0m segments, what magnetic force does each segment exert on the other? (answer in F= ? N)

Part B.) Is the force found in part (A) attractive or repulsive?

Answer: 0.024 N attraction force is being applied.

Explanation:

See attached picture.

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22. reduction
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3 years ago
A mechanic pushes a 2.30 ✕ 103-kg car from rest to a speed of v, doing 4,800 J of work in the process. During this time, the car
Illusion [34]

The horizontal force applied is  160 N while the velocity is  2.03 m/s.

<h3>What is the speed of the car?</h3>

The work done by the car is obtained as the product of the force and the distance;

W = F x

F = ?

x = 30.0 m

W = 4,800 J

F = 4,800 J/30.0 m

F = 160 N

But F = ma

a = F/m

a = 160 N/2.30 ✕ 10^3-kg

a= 0.069 m/s

Now;

v^2 = u^2 + 2as

u = 0/ms because the car started from rest

v = √2as

v = √2 * 0.069 * 30

v = 2.03 m/s

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5 0
2 years ago
An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99
sashaice [31]

<u>Answer:</u> The average atomic mass of the given element is 20.169 amu.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of the isotopes each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i     .....(1)

We are given:

  • For isotope 1:

Mass of isotope 1 = 19.99 amu

Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

  • For isotope 2:

Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26%

Fractional abundance of isotope 2 = 0.0026

  • For isotope 3:

Mass of isotope 3 = 21.99 amu

Percentage abundance of isotope 3 = 8.82%

Fractional abundance of isotope 3 = 0.0882  

Putting values in equation 1, we get:

\text{Average atomic mass}=[(19.99\times 0.9092)+(20.99\times 0.0026)+(21.99\times 0.0882)]

\text{Average atomic mass}=20.169amu

Hence, the average atomic mass of the given element is 20.169 amu.

4 0
3 years ago
PLEASE HELP FAST The object distance for a convex lens is 15.0 cm, and the image distance is 5.0 cm. The height of the object is
IgorLugansk [536]

Answer:

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Explanation:

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image distance, d_i = 5.0 cm

height of the object, h_o = 9.0 cm

height of the image, h_i = ?

Apply lens equation;

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How did life evolve from nonliving matter?
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Answer: it  is proposes that in Earth's prebiotic history, simple organic matter was exposed to energy in the form of volcanoes and electrical storms

Explanation:

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