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3 years ago

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A large pendulum with a 200-lb gold plated bob 12 inches in diameter is on display in the lobby of the United Nations building.

The pendulum has a length of 75 ft. It is used to show the rotation of the Earth—for this reason it is referred to as a Foucault pendulum.Part A) If the pendulum were to be taken to the Moon, where the acceleration of gravity is g/6, would the period increase, decrease, or stay the same?
Part B) Check your result in part A by calculating the period of the pendulum on the Moon.

1 answer:

xeze [42]3 years ago

3 0

Answer:

9.59143 s

23.4941 s

Explanation:

L = Length of pendulum = 75 ft

g = Acceleration due to gravity = 9.81 m/s²

Time period is given by

$T=2\pi \sqrt{\dfrac{L}{g}}$

On Earth

$T=2\pi \sqrt{\dfrac{75\times 0.3048}{9.81}}\\\Rightarrow T=9.59143\ s$

The time period on Earth is 9.59143 s

On Moon

$T=2\pi \sqrt{\dfrac{75\times 0.3048}{\dfrac{9.81}{6}}}\\\Rightarrow T=23.4941\ s$

The time period on Moon is 23.4941 s

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9* $10^{-5}$ m

Explanation:

Step 1:

We are given the initial length of the Pyrex glass dish at a particular temperature and need to calculate the change in the length when the temperature changes by 120° C. The coefficient of linear expansion of Pyrex is provided.

Step 2:

Change in length = Coefficient of linear expansion * Change in temperature * Initial length

Step 3:

Coefficient of linear expansion = 3* $10^{-6}$ /°C

Change in temperature = 120°C = 120 K

Initial length = 0.25 m

Step 4:

Change in length = 3* $10^{-6}$ * 120 * 0.25 = 9* $10^{-5}$ m

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Read 2 more answers

39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th

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<u>Answer:</u> The final temperature of the solution is $80.14^oC$

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, $\text{heat}_{absorbed}=\text{heat}_{released}$

To calculate the amount of heat released or absorbed, we use the equation:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

Also,

$m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]$ ..........(1)

where,

q = heat absorbed or released

$m_1$ = mass of aluminium = 39 g

$m_2$ = mass of coffee = 166 g

$T_{final}$ = final temperature = ?

$T_1$ = temperature of aluminium = $24^oC$

$T_2$ = temperature of coffee = $83^oC$

$c_1$ = specific heat of aluminium = $0.904J/g^oC$

$c_2$ = specific heat of coffee= $4.1801J/g^oC$

Putting all the values in equation 1, we get:

$39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]$

$T_{final}=80.14^oC$

Hence, the final temperature of the solution is $80.14^oC$

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