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ankoles [38]
2 years ago
15

In pottery class, you throw a pot from a lump of wet clay. your pot's mass is 5.5 kg. after the pot is fired, it's mass is 4.9 k

g. the density of wet clay is 1.60 g/cm3 and the density of fired clay is 1.36 g/cm3. what was the volume of your pot before it was fired? what was the volume of the pot after it was fired? express your answer in significant digits.
Physics
1 answer:
kondor19780726 [428]2 years ago
8 0

We can calculate for the volume by taking the ratio of mass and density, that is:

volume = mass / density

 

A. when mass = 5.5 kg = 5500 g; density = 1.60 g/cm^3

volume = 5500 g / (1.60 g/cm^3)

volume = 3,437.5 cm^3

Taking the significant digits:

volume = 3,400 cm^3 = 3.4 L

 

B. when mass = 4.9 kg = 4900 g; density = 1.36 g/cm^3

volume = 4900 g / (1.36 g/cm^3)

volume = 3,602.94 cm^3

Taking the significant digits:

<span>volume = 3,600 cm^3 = 3.6 L</span>

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Refer to the diagram shown below.

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The total displacement for the first leg of the trip is
\vec{d} = \vec{dN} + \vec{dW} + \vec{dS} \\ \vec{d}= 1.0\hat{j}-0.6\hat{i}-0.2\hat{j} \\ \vec{d}=-0.6\hat{i}+0.8\hat{j}

Answer:
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Consider a bird that flies at an average speed of 10.7 m/sm/s and releases energy from its body fat reserves at an average rate
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t=\dfrac{E}{P}\\\Rightarrow t=\dfrac{157393.6}{3.7}\\\Rightarrow t=42538.811\ s

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s=vt\\\Rightarrow s=10.7\times 42538.811\\\Rightarrow s=455165.2777\ m

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At time t, gives the position of a 3.0 kg particle relative to the origin of an xy coordinate system ( ModifyingAbove r With rig
Elena-2011 [213]

Complete Question

  The complete Question is shown on the first uploaded image

Answer:

a

The torque acting on the particle is  \tau = 48t \r k

b

The magnitude of the angular momentum increases relative to the origin

Explanation:

From the equation we are told that

      The position of the particle is   \= r = 4.0 t^2 \r i - (2.0 t - 6.0 t^2 ) \r j

       The mass of the particle is m = 3.0 kg

        The time is  t

   

The torque acting on  the particle is mathematically represented as

           \tau = \frac{ d \r l }{dt}

where \r l is change in angular momentum which is mathematically represented as

       \r l = m (\r r \ \ X  \ \ \r v)

Where X mean cross- product

   \r v is the velocity which is mathematically represented as

           \r v = \frac{d \r r }{dt}

Substituting for  \r r

           \r v = \frac{d }{dt} [ 4 t^2 \r i - (2t + 6t^2 ) \r j]

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          \r r  \  \ X \ \ \r v = \left[\begin{array}{ccc}{\r i}&{\r j}&{\r k}\\{4t^2}&{-2t -6t^2}&0\\{8t}&{-2 -12t}&0\end{array}\right]

                       = 0 \r i + 0 \r j + (- 8t^2 -48t^3 + 16t^2 + 48t^3 ) \r k

                      \r r \ \  X \ \ \r v = 8t^2 \r k

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       \tau = 48t \r k

The magnitude of the angular momentum can be evaluated mathematically as

        |\r l| = \sqrt{(24 t^2) ^2}

        |\r l| = 24 t^2

From the is equation we see that the magnitude of the angular momentum is varies directly with square of the time so it would relative to the origin because at the origin t= 0s and we move out from origin t increases hence angular momentum increases also

4 0
2 years ago
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