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liberstina [14]
2 years ago
6

Compare the weight of a mountain climber when she is at the bottom of a mountain with her weight when she is at the top of the m

ountain. In which case is her weight larger? a. Both are the same. b. She weighs four times as much at the top. c. She weighs twice as much at the top. d. She weighs more at the top. e. She weighs more at the bottom.
Physics
1 answer:
kakasveta [241]2 years ago
4 0

Answer:

The correct answer is  a. Both are the same

Explanation:

For this calculation we must use the gravitational attraction equation

    F = G m M / r²

Where M will use the mass of the Earth, m the mass of the girl and r is the distance of the girl to the center of the earth that we consider spherical

To better visualize things, let's repair the equation a little

     F = m (G M / r²)

The amount in parentheses called acceleration of gravity, entered the force called peos

     g = G M / r²

     F = W

    W = m g

When analyzing this equation we see that the variation in the weight of the girl depends on the distance, which is the radius of the earth plus the height where the girl is

    r = Re + h

    Re = 6.37 10⁶ m

    r² = (Re + h)²

    r² = Re² (1 + h / Re)²

Let's replace

    W = m (GM / Re²)   (1+ h / Re)⁻²

    W = m g   (1+ h / Re)⁻²

This is the exact expression for weight change with height, but let's look at its values ​​for some reasonable heights h = 6300 m (very high mountain)

     h / Re = 10 ⁻³

     (1+ h / Re)⁻² = 0.999⁻²

Therefore, the negligible weight reduction, therefore, for practical purposes the weight does not change with the height of the mountain on Earth

The correct answer is a

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1. What is the new kinetic energy of the 1900kg ship on the right moving at 4 m/s?
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6 0
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A 300 g ball and a 600 g ball are connected by a 40-cm-lon massless, rigid rod. The structure rotates about its center of me at
Readme [11.4K]

Answer:

 KE = 1.75 J

Explanation:

given,

mass of ball, m₁ = 300 g = 0.3 Kg

mass of ball 2, m₂ = 600 g = 0.6 Kg

length of the rod = 40 cm = 0.4 m

Angular speed = 100 rpm= 100\times \dfrac{2\pi}{60}

                         =10.47\ rad/s

now, finding the position of center of mass of the system

    r₁ + r₂ = 0.4 m.....(1)

 equating momentum about center of mass

  m₁r₁ = m₂ r₂

   0.3 x r₁ = 0.6 r₂

   r₁ = 2 r₂

Putting value in equation 1

2 r₂ + r₂ = 0.4

 r₂ = 0.4/3

 r₁ = 0.8/3

now, calculation of rotational energy

KE = \dfrac{1}{2}I_1\omega^2+\dfrac{1}{2}I_2\omega^2

KE = \dfrac{1}{2}\omega^2 (I_1 +I_2)

KE = \dfrac{1}{2}\omega^2 (m_1r_1^2 +m_2r^2_2)

KE = \dfrac{1}{2}\times 10.47^2(0.3\times (0.8/3)^2 +0.6\times (0.4/3)^2)

 KE = 1.75 J

the rotational kinetic energy is equal to 1.75 J

7 0
3 years ago
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