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2 years ago

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A 2.00 kg cat is in a 97.00 kg elevator. What force on the elevator cable would be needed to lower the cat/elevator pair with an

acceleration of 2.00 m/s^2 downwards?

1 answer:

umka21 [38]2 years ago

3 0

The magnitude of force on the elevator cable that would be needed to lower the cat/elevator pair is 198 Newton.

<u>Given the following data:</u>

Mass of cat = 2 kg

Mass of elevator = 97 kg

Acceleration = 2 $m/s^2$

To determine the magnitude of force on the elevator cable that would be needed to lower the cat/elevator pair, we would apply Newton's Second Law of Motion:

First of all, we would calculate the total mass of the cat/elevator pair.

$Total \;mass=2 + 97$

Total mass = 99 kilograms

Mathematically, Newton's Second Law of Motion is given by this formula;

$Force = mass \times acceleration$

Substituting the given parameters into the formula, we have;

$Force = 99 \times 2$

Net force = 198 Newton

Read more here: brainly.com/question/24029674

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Answer:

1) Option E is correct.

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2) Option C is correct.

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4) Option B is correct.

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Explanation:

1) vector v has initial point (8,1) and terminal point (6,4)

Write vector v as a linear combination of the standard unit vector.

v represented in standard form is given as

Vector v = (final position vector) - (initial position vector) = (6î + 4ĵ) - (8î + ĵ) = (-2î + 3ĵ)

2) v be the vector with initial point (−5,1) and terminal point (5,3). Find the vertical component of this vector.

v represented in standard form is given as

Vector v = (final position vector) - (initial position vector) = (5î + 3ĵ) - (-5î + ĵ)

v = (10î + 2ĵ)

The vertical component is the ĵ-component and it is equal to 2

3) Find the sum of the vectors u =6i −4j and v⃗ =−2i −5j

Vector sum is done on a per component basis

Sum = u + v = (6î - 4ĵ) + (-2î - 5ĵ) = (4î - 9ĵ)

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u = (-2î - 3ĵ)

v = (5î + 2ĵ)

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Magnitude of a vector is given as

/v/ = √[vₓ² + vᵧ²]

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/v/ = √[(4²) + (-4)²] = 4√2 units = 5.66 units

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Unit vector in the direction of a vector = (vector)/(magnitude of vector)

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Answer:

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(d). The maximum magnitude of electric field is $1.54\times10^{4}\ N/C$

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