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3 years ago

An 80kg student running at 3.50m/s grabs a rope that is hanging vertically. How high will the student swing?

Physics

1 answer:

Rudiy273 years ago

8 0

Answer:

The student will swing to a height of 61.25 cm.

Explanation:

Given that,

Mass of the student, m = 80 kg

Speed of student, v = 3.5 m/s

The student running at this speed grabs a rope that is hanging vertically. In the whole process, the energy of the system remains conserved. So, using conservation of energy to find the height reached by the student. So,

$\dfrac{1}{2}mv^2=mgh$

h is height reached by the student

$h=\dfrac{v^2}{2g}\\\\h=\dfrac{(3.5)^2}{2\times 10}\\\\h=0.6125\ m\\\\h=61.25\ cm$

So, the student will swing to a height of 61.25 cm.

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Answer:

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3 years ago

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3 years ago

Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms f

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Answer:

Her angular speed (in rev/s) when her arms and one leg open outward is 1.4 rev/s

Explanation:

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3 years ago

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Answer:

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3 years ago

A 57 kg boy and a 41 kg girl use an elastic rope while engaged in a tug-of-war on a frictionless icy surface. If the acceleratio

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Answer:

Acceleration of the boy a₁:

$a_{1} = 1.87 \frac{m}{s^{2} }$

Explanation:

Conceptual analysis

We apply Newton's second law to the boy and the girl:

F = m*a (Formula 1)

F : Force in Newtons (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

Nomenclature

m₁ : boy mass

m₂ : girl mass

a₁ : boy acceleration

a₂ : girl acceleration

F₁ : boy acceleration

F₂ : girl acceleration

Known data

m₁ = 57 kg

m₂ = 41 kg

a₂ = 2.6 m/s²

Problem development

We apply to Newton's third law of action and reaction, then:

F₁ = F₂ , We apply the formula (1):

m₁*a₁ = m₂*a₂

$a_{1} = \frac{m_{2}* a_{2} }{m_{1} }$

$a_{1} = \frac{41* 2.6 }{ 57 }$

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5 0

3 years ago