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k0ka [10]
3 years ago
11

A standard 1 kilogram weight is a cylinder 54.0 mm in height and 55.0 mm in diameter. what is the density of the material

Physics
1 answer:
denis-greek [22]3 years ago
5 0

The radius of the cylinder is equal to half the diameter:

r=\frac{d}{2}=\frac{55.0 mm}{2}=27.5 mm

The volume of the cylinder is given by:

V=\pi r^2 h=\pi (27.5 mm)^2 (54.0 mm)=1.28 \cdot 10^5 mm^3

where h is the heigth of the cylinder. Converting into meters,

V=1.28 \cdot 10^{-4} m^3

And the density of the material will be given by the ratio between the mass and the volume:

d=\frac{m}{V}=\frac{1 kg}{1.28 \cdot 10^{-4} m^3}=7812.5 kg/m^3

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Please answer this question for me and explain why.
horsena [70]

Answer:

D.None of these

Explanation:

The derivation of acceleration formula:

Let us call the 5kg mass m_2 and the 4kg mass m_1. If the tension in the string is T then for the mass m_2

(1). T-m_2g=-m_2a <em>(the negative sign on the right side indicates that acceleration is downwards)</em>

And for the mass m_1

(2). T-m_1g =m_1a<em> (the acceleration is upwards, hence the positive sign)</em>

Solving for T in the 2nd equation we get:

T =m_1a+m_1g,

and putting this into the 1st equation we get:

m_1a+m_1g-m_2g=-m_2a\\\\m_1a+m_2a = m_2g-m_1g\\\\a(m_1+m_2)= (m_2-m_1)g

\boxed{a= \dfrac{(m_2-m_1)}{(m_1+m_2)} g}

Back to the question:

Using the formula for the acceleration we find

a= \dfrac{(5kg-4kg)}{(5kg+4kg)} g

a = \dfrac{g}{9},

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7 0
3 years ago
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Answer:

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Working in SI units, Power = Watts, Work = Joules, Time = seconds.

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