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andreyandreev [35.5K]
3 years ago
13

Una mujer de masa m está parada en el borde de una mesa giratoria horizontal de momento de inercia I y radio R. La mesa al princ

ipio está en reposo y tiene libertad de moverse alrededor de un eje vertical sin fricción que pasa por su centro. La mujer empieza a caminar alrededor de la orilla en direción de las manecillas del reloj (vista desde arriba) a una velocidad constante v en relación a la Tierra.
a) En qué dirección y con qué velocidad angular gira la mesa?

b) Cuál es el trabajo efectuado por la mujer para poner en movimiento la mesa?
Physics
1 answer:
Dovator [93]3 years ago
3 0

r

- \frac{mR^2 }{I  } \ vAnswer:

a)      w = - \frac{m r }{I} v  ,  b)   W = - ½ m_woman R² (1 + m_woman R / I²) v²

Explanation:

a) To solve this exercise, let's use the conservation of angular momentum.

We define a system formed by the table and the woman, therefore the torques are internal and the moment is conserved

initial instant. Before starting to move the woman

         L₀ = 0

final instant. After starting to move

         L_f = I w + m v r

the moment is preserved

        L₀ = L_f

         0 = Iw + m v r

         w = - \frac{m r }{I} v                    (1)

the direction of the angular velocity is opposite to the direction of the linear velocity, that is, counterclockwise

b) for this part we use the relationship between work and kinetic energy

        W = ΔK

in this case the initial speed is zero and the final speed of the table, using the relationship between linear and angular variables

         v = w r

we substitute

          W = 0 - ½ I_total w²

          I_total = I + m_{woman} R²

          W = - ½ (I + m_woman R²)  ( \frac{m_{woman} R}{I} \ v) ²

          W = - ½ (m_woman² R² + m_woman³ R³ / I²) v²

          W = - ½ m_woman R² (1 + m_woman R / I²) v²

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The table below shows different items that are found sitting on the counter in a kitchen. The temperature of the kitchen is 72 d
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This question is incomplete

Complete Question

The table below shows different items that are found sitting on the counter in a kitchen. The temperature of the kitchen is 72 degrees Fahrenheit. Classify the objects according to whether their molecules will speed up or slow down after being left in the kitchen for a period of time.

Drag the objects to the correct category.

Object Temperature (Fahrenheit)

ice cubes 26°F

glass of tea 60°F

cooked piece of meat 160°F

butter 55°F

pot of water 75°F

bowl of soup 140°F

Answer:

Molecules that would speed up

ice cubes 26°F

glass of tea 60°F

butter 55°F

Molecules that would slow down

cooked piece of meat 160°F

pot of water 75°F

bowl of soup 140°F

Explanation:

Molecules that are found or contained in a substances have the tendency to react in such as what that they are are sped up or slowed down. This is due to their exposure to various kinds of changes in temperature, which could be a hot temperature , cold temperature e.t.c

When a substance is exposed to a hot temperature, the molecules of that substance speed up depending on how hot the temperature is while when a substance is cooled down or exposed to cold temperature, the molecules tend to slow down.

In the question above, the temperature of the kitchen is 72 degrees Fahrenheit.

Molecules that would speed up after been left in the Kitchen for a while:

a) Ice cubes 26°F: This is because 72°F is a warm took temperature, so the ice cubes would melt causing the molecules to speed up.

b)Glass of tea 60°F: This glass of tea is at a cool temperature of 60°F, when it is kept in the Kitchen which has a temperature of 72°F for while, the temperature of the glass of tea would increase due to an increase in its rate of reaction causing the molecules of the glass of water to speed up.

c) Butter 55°F: This is because 72°F is a warm temperature, so the butter would melt, increasing its rate of reaction and causing the molecules to speed up

Molecules that would slow down

a) Cooked piece of meat 160°F : 160°F is a very hot temperature and when it is left in a the Kitchen(72°F) , the cooked meat would begin to cool down and drop in temperature from 160°F, causing the molecules to slow down.

b) Pot of water 75°F: The temperature of the kitchen and the temperature of the pot of water is the same, hence the molecules of the water would slow down.

c) Bowl of soup 140°F : This bowl of soup is very hot and when it is left in the Kitchen(72°F) , the bowl of soup would begin to cool down and drop in temperature from 140°F, slowing down the rate of reaction and causing the molecules to slow down.

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