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Ksivusya [100]
3 years ago
8

Is it c or its not??

Mathematics
2 answers:
Bingel [31]3 years ago
6 0

Answer:

C)  0.9

Step-by-step explanation:

density = mass / volume

density = 82 / 96

density = 0.9 g/mL

morpeh [17]3 years ago
6 0

Answer: Choice C) 0.9

You are correct.

==========================================================

Explanation:

1 cc = 1 cubic centimeter = 1 milliliter = 1 mL

Let's find the density of each ball. To do so, divide the mass over the volume

  • Density of Ball X = mass/volume = (25 g)/(24 ml) = 1.042 g/mL approx
  • Density of Ball Y = mass/volume = (82 g)/(96 ml) = 0.854 g/mL approx
  • Density of Ball Z = mass/volume = (14.2 g)/(14.8 ml) = 0.959 g/mL approx

We're told that Ball X and Ball Z sink, while Ball Y floats. This must mean the density of the liquid is somewhere between 0.854 g/mL and 0.959 g/mL.

Of the four answer choices, only choice C has its value between 0.854 g/mL and 0.959 g/mL. So you are correct.

------------------------

Extra info:

If the density of the liquid was less than 0.854 g/mL, then Ball Y would sink. Recall that objects only float in liquid if the object's density is less than that of the liquid's density. The more dense stuff sinks to the bottom.

If the density of the liquid was greater than 0.959 g/mL, then both Ball Y and Ball Z would float (Ball X would only float if the liquid density is larger than 1.042 g/mL)

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mars1129 [50]

Answer:

y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+\frac{1}{\sqrt{3}}\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Step-by-step explanation:

A second order linear , homogeneous ordinary differential equation has form ay''+by'+cy=0.

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Let y=e^{rt} be it's solution.

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\left ( r^2+r+1 \right )e^{rt}=0

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{ we know that for equation ax^2+bx+c=0, roots are of form x=\frac{-b\pm \sqrt{b^2-4ac}}{2a} }

We get,

y=\frac{-1\pm \sqrt{1^2-4}}{2}=\frac{-1\pm \sqrt{3}i}{2}

For two complex roots r_1=\alpha +i\beta \,,\,r_2=\alpha -i\beta, the general solution is of form y=e^{\alpha t}\left ( c_1\cos \beta t+c_2\sin \beta t \right )

i.e y=e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

Applying conditions y(0)=1 on e^{\frac{-t}{2}}\left ( c_1\cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right ), c_1=1

So, equation becomes y=e^{\frac{-t}{2}}\left ( \cos\left ( \frac{\sqrt{3}t}{2} \right )+c_2\sin \left ( \frac{\sqrt{3}t}{2} \right ) \right )

On differentiating with respect to t, we get

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