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monitta
1 year ago
11

Explain how you would find the volume of a regular-shaped cube in comparison to an irregular shaped object such as a rock.

Physics
1 answer:
Colt1911 [192]1 year ago
4 0

1. To obtain the volume of a regular-shaped cube, we simply use the formula.

2. To find the volume of an irregular shaped object, we simply use displace method

<h3>Volume of regular shape cube</h3>

The volume of a cube can be obtained by using the formula below:

Volume = Length × Width × Height

<h3>Volume of irregular object</h3>

The volume of irregular objects can be obtained by using displacement method as illustrated below:

  • Volume of water = 20 mL
  • Volume of water + stone = 25 mL
  • Volume of stone =?

Volume of stone = (Volume of water + stone) - (volume of water)

Volume of stone = 25 - 20

Volume of stone = 5 mL

Thus, we can see that the volume of regular objects can be obtained by using their formulas whereas, the volume of irregular objects can be obtained by using displacement method.

Learn more about volume of irregular shape:

brainly.com/question/22851494

#SPJ1

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3 years ago
A building made with a steel structure is 565 m high on a winter day when the temperature is 0◦F. How much taller is the buildin
anyanavicka [17]

To solve this problem we apply the thermodynamic equations of linear expansion in bodies.

Mathematically the change in the length of a body is subject to the mathematical expression

\Delta L = L_0 \alpha \Delta T

Where,

L_0 = Initial Length

\alpha = Thermal expansion coefficient

\Delta T = Change in temperature

Since we have values in different units we proceed to transform the temperature to degrees Celsius so

0\°F \Rightarrow (0-32)*\frac{5}{9} = -17.77\°C

103\°F \Rightarrow (103-32)*(\frac{5}{9})= 39.44\°C

The coefficient of thermal expansion given is

\alpha = 1.1*10^{-5}/\°C

The initial length would be,

L_0 = 565m

Replacing we have to,

\Delta L = L_0 \alpha \Delta T

\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))

\Delta L = (565)(1.1*10^{-5})(39.44-(-17.77))

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3 years ago
A disk is uniformly accelerated from rest with angular acceleration α. The magnitude of the linear acceleration of a point on th
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Answer:

a = R\alpha\sqrt{1 + \alpha^2t^4}

Explanation:

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a_t = R\alpha

2) Centripetal acceleration

a_c = \omega^2 R

here we know that

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a_c = (\alpha t)^2 R

now we know that net linear acceleration is given as

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so we have

a = \sqrt{R^2\alpha^2 + R^2(\alpha t)^4}

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