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soldier1979 [14.2K]
3 years ago
5

Adita lifts a book from the floor, carries it across the room, and places it on a high shelf. When is Adita doing work on the bo

ok?
She does work on the book at no time during this process.
She does work only while she carries the book across the room.
She does work from the moment she touches the book until she lets it go.
She does work as she picks up the book and as she lifts it to the shelf.
Physics
2 answers:
noname [10]3 years ago
5 0

She does work from the moment she touches the book until she lets it go. Work is anything that requires energy. Therefore, she is working as she picks up the book, carries it, and when she is lifting it onto the shelf.

ZanzabumX [31]3 years ago
3 0

option d

took test on edu

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Which is the graph of the polar equation r = 4 cos3 theta
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Im pretty sure it’s d

Explanation:

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A billiard ball collides with a second identical ball in an elastic head-on collision. What is the kinetic energy of the system
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Answer:

Explanation:

From the question, we were made to understand that the collision between the two billiard balls was an elastic collision. Hence, an elastic collision is one in which the kinetic energy is conserved. Meaning the kinetic energy before the collision is still retained after the collision.

Kinetic energy before collision = kinetic energy after collision

1/2mv^2 = 1/2mv^2

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80 grams of iron at 100°C is dropped into 200 of water at 20°C contained in an iron vessel of mass 50 gram find the resulting te
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Answer:

the resulting temperature is 23.37 ⁰C

Explanation:

Given;

mass of the iron, m₁ = 80 g = 0.08 kg

mass of the water, m₂ = 200 g = 0.2 kg

mass of the iron vessel, m₃ = 50 g = 0.05 kg

initial temperature of the iron, t₁ = 100 ⁰C

initial temperature of the water, t₂ = 20 ⁰C

specific heat capacity of iron, c₁ = 462 J/kg⁰C

specific heat capacity of water, c₂ = 4,200 J/kg⁰C

let the temperature of the resulting mixture = T

Apply the principle of conservation of energy;

heat lost by the hot iron = heat gained by the water

m_1c_1 \Delta t_1 = m_2c_2\Delta t_2\\\\m_1c_1 (100 - T) = m_2c_2 (T- 20)\\\\0.08 \times 462 (100-T) = 0.2 \times 4,200 (T-20)\\\\36.96 (100-T) = 840 (T-20) \\\\100 - T = 22.72 (T-20)\\\\100-T = 22.72 T - 454.4 \\\\554.4 = 23.72T\\\\T = \frac{554.4}{23.72} \\\\T = 23.37 \ ^0C

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6 0
3 years ago
A 10-foot ladder is leaning straight up against a wall when a person begins pulling the base of the ladder away from the wall at
docker41 [41]

Answer:

y = 4.36

Explanation:

Let the height of the ladder be L

L = 10

Also:

  • Let x = distance\ from\ the\ base\ of\ the\ ladder
  • Let y = height\ of\ the\ base\ of\ the\ ladder

When the ladder leans against the wall, it forms a triangle and the length of the ladder forms the hypotenuse.

So, we have:

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When the base is 9ft from the wall, this means that:

x = 9

Substitute 9 for x and 10 for L in L^2 = x^2 + y^2

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Make y^2 the subject

y^2 = 100 - 81

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Make y the subject

y = \sqrt{19

y = 4.36

<em>Hence, the true distance at that point is approximately 4.36ft</em>

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