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3 years ago

12

A wrecking ball is suspended by two cables as shown below. If the tension in cable 2 is 12 000 N, what is the weight of the wrec

king ball?

1 answer:

Paul [167]3 years ago

5 0

Wrecking ball wow get Miley Cyrus also I’m sorry I’m joking bout it and you should report the person who be putting “links”

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Most automobiles have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made o

Colt1911 [192]

Answer:

There is a loss of fluid in the container of 0.475L

Explanation:

To solve the problem it is necessary to take into account the concepts related to the change of voumen in a substance depending on the temperature.

The formula that describes this thermal expansion process is given by:

$\Delta V = \beta V_0 \Delta T$

Where,

$\Delta V =$ Change in volume

$V_0 =$ Initial Volume

$\Delta T =$ Change in temperature

$\beta =$ coefficient of volume expansion (Coefficient of copper and of the liquid for this case)

There are two types of materials in the container, liquid and copper, so we have to change the amount of Total Volume that would be subject to,

$\Delta V_T = \Delta V_l - \Delta V_c$

Where,

$\Delta V_l$ = Change in the volume of liquid

$\Delta V_c$ = Change in the volume of copper

Then replacing with the previous equation we have:

$\Delta V = \beta_l V_0 \Delta T- \beta_c V_0 \Delta T$

$\Delta V = (\beta_l-\beta_c)V_0\Delta T$

Our values are given as,

Thermal expansion coefficient for copper and the liquid to 20°C is

$\beta_c = 51*10^{-6}/\°C$

$\beta_l = 400*10^{-6}/\°C$

$V_0 = 16L$

$\Delta T = (95\°C-10\°C)$

Replacing we have that,

$\Delta V = (\beta_l-\beta_c)V_0\Delta T$

$\Delta V = (400*10^{-6}/\°C-51*10^{-6}/\°C)(16L)(95\°C-10\°C)$

$\Delta V = 0.475L$

Therefore there is a loss of fluid in the container of 0.475L

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