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muminat
3 years ago
9

Two cylinders are made of the same material. Cylinder A is one-fourth (1/4) the length of cylinder B and it has a radius that is

four times greater than the radius of cylinder B. What is the ratio of the mass of cylinder A to the mass of cylinder B?
Chemistry
2 answers:
tankabanditka [31]3 years ago
8 0

Answer:

M_{A} : M_{B}  = 4 : 1

Explanation:

Given that:

Volume of a cylinder = \pir^{2}h

For cylinder A, l_{A} = h_{A} = \frac{1}{4} l_{B} and r_{A} = 4r_{B}.

Volume of cylinder A = \pi(4r_{B}) ^{2} × \frac{1}{4} l_{B}

                                   = 4\pir_{B} ^{2} l_{B}

Volume of cylinder B =  \pi(r_{B}) ^{2} l_{B}

                                  =  \pir_{B} ^{2}l_{B}

To determine the ratio of their masses, density (ρ) is defined as the ratio of the mass (M) of a substance to its volume (V).

i.e    ρ = \frac{M}{V}

Thus, since the cylinders are made from the same material, they have the same density  (ρ). So that;

density of A = density of B

density of A = \frac{M_{A} }{4\pi r_{B} ^{2}l_{B}    }

density of B = \frac{M_{B} }{\pi r_{B} ^{2}l_{B}  }

⇒            \frac{M_{A} }{4\pi r_{B} ^{2}l_{B}    }  =  \frac{M_{B} }{\pi r_{B} ^{2}l_{B}  }

The ratio of mass of cylinder A to that of B is given as;

                       \frac{M_{A} }{M_{B} }  =  \frac{4\pi r_{B} ^{2} l_{B}  }{\pi r_{B} ^{2} l_{B}  }

⇒                        \frac{M_{A} }{M_{B} } = \frac{4}{1}

Therefore, M_{A} : M_{B}  = 4 : 1

Kruka [31]3 years ago
3 0

Answer:

4.

Explanation:

Hello,

In this case, since we are talking about the same material, their densities are the same:

\rho _A=\rho _B

And each density is defined by:

\rho _A=\frac{m_A}{V_A} \\\\\rho _B=\frac{m_B}{V_B}

Thus, we also define the volume of a cylinder:

V_{cylinder}=\pi r^2h

Therefore, we obtain:

\rho _A=\frac{m_A}{\pi r_A^2h_A}

\rho _B=\frac{m_B}{ \pi r_B^2h_B}

Now, the given information regarding the the length and the radius is written mathematically:

h_A=\frac{1}{4} h_B\\\\r_A=4 r_B

So we introduce such additional equations in:

\frac{m_A}{\pi r_A^2h_A}=\frac{m_B}{\pi r_B^2h_B}\\\\\frac{m_A}{\pi (4r_B)^2(\frac{1}{4}h_B)}=\frac{m_B}{\pi r_B^2h_B}\\\\\frac{m_A}{m_B} =\frac{\pi (4r_B)^2(\frac{1}{4}h_B)}{\pi r_B^2h_B}

So we simplify for the radius and lengths:

\frac{m_A}{m_B} =\frac{\pi (4r_B)^2(\frac{1}{4}h_B)}{\pi r_B^2h_B}\\\\\frac{m_A}{m_B} =16 *\frac{1}{4}\\ \\\frac{m_A}{m_B} =4

So the ratio of the mass of cylinder A to the mass of cylinder B is 4.

Best regards.

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