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3 years ago

Physics question about density

Physics

2 answers:

scZoUnD [109]3 years ago

7 0

So the right answer is 4kg/m^3.

Look at the attached picture

Hope it will help you

Good luck on your assignment

ohaa [14]3 years ago

6 0

Answer:

The answer is option 4.

Explanation:

Firstly, you have to convert the mass of tea bagto kilogram as the question wants the unit in kg/m³.

$1000g = 1kg$

$100g = 1kg \div 10$

$100g = 0.1kg$

So the mass of tea bag is 0.1 kg. Next you have to use density formula ρ = mass/volume. Then you have to substitute the values into the formula :

$ρ = \frac{mass}{volume}$

Let mass = 0.1kg,

Let volume = 0.025m³,

$ρ = \frac{0.1}{0.025}$

$ρ = 4kg \: per \: {m}^{3}$

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Jobisdone [24]

Answer:

The tension is $T= \frac{11}{2\sqrt{3} } Mg$

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Explanation:

From the question we are told that

The mass of the beam is $m_b =M$

The length of the beam is $l = L$

The hanging mass is $m_h = 3M$

The length of the hannging mass is $l_h = \frac{3}{4} l$

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The free body diagram of this setup is shown on the first uploaded image

The force $F_x \ \ and \ \ F_y$ are the forces experienced by the beam due to the hinges

Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

$\sum F =0$

Now about the x-axis the moment is

$F_x -T cos \theta = 0$

=> $F_x = Tcos \theta$

Substituting values

$F_x =T cos (60)$

$F_x= \frac{T}{2} ---(1)$

Now about the y-axis the moment is

$F_y + Tsin \theta = M *g + 3M *g ----(2)$

Now the torque on the system is zero because their is no rotation

So the torque above point 0 is

$M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0$

$\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0$

$\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}$

$T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }$

$T= \frac{11}{2\sqrt{3} } Mg$

The horizontal force provided by the hinge is

$F_x= \frac{T}{2} ---(1)$

Now substituting for T

$F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}$

$Fx= \frac{11}{4\sqrt{3} } Mg$

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