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klasskru [66]
3 years ago
6

Bird bones have air pockets in them to reduce their weight–this also gives them an average density significantly less than that

of the bones of other animals. suppose an ornithologist weighs a bird bone in air and in water and finds its mass is 43.0 g and its apparent mass when submerged is 3.60 g (the bone is watertight).a. what mass of water is displaced? b. what is the volume of the bone? c. what is its average density?
Physics
1 answer:
Ksenya-84 [330]3 years ago
7 0

Answer:

39.4 g

39.4 cm³

1.09137 g/cm³

Explanation:

\rho = Density of water = 1 g/cm³

Mass of water displaced will be the difference of the

m=43-3.6\\\Rightarrow m=39.4\ g

Mass of water displaced is 39.4 g

Density is given by

\rho=\dfrac{m}{v}\\\Rightarrow v=\dfrac{m}{\rho}\\\Rightarrow v=\dfrac{39.4}{1}\\\Rightarrow v=39.4\ cm^3

So, volume of bone is 39.4 cm³

Average density of the bird is given by

\rho=\dfrac{43}{39.4}\\\Rightarrow \rho=1.09137\ g/cm^3

The average density is 1.09137 g/cm³

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A box of mass 64 kg is at rest on a horizontal frictionless surface. A constant horizontal force F~ then acts on the box and acc
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Answer:

The correct answer is;

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3 0
3 years ago
A slab of insulating material has thickness 2d and is oriented so that its faces are parallel to the yz-plane and given by the p
olganol [36]

Answer:

a) In the center of the slab there are no charges inside, so by Gauss's law the electric field is zero,

b)  E = ρ x / 2ε₀   , c)   E = σ / 2ε₀,

d) the direction of the electric field as the charge is positive is leaving the plate  

Explanation:

a) In the center of the slab there are no charges inside, so by Gauss's law the electric field is zero,

Another way of analyzing it is that the charge on one side of the crockery creates an outgoing electric field in the center, the charge on the other side of the crockery creates a field of equal magnitude, but in the opposite direction, so the resulting field is zero. .

b) Let's use Gauss's law to calculate the electric field, let's use as cylinder a Gaussian surface with the base parallel to the faience, so the scalar product is reduced to the algebraic product

         Ф = ∫ E. dA = qint /ε₀

 

The slab area is A

let's use the concept of charge density

         ρ = qint / V

the volume of the slab is the area times the thickness

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       qint = ρ A x

as the two sides of the slab create an electric field the flow is

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where x goes from zero to the thickness of the plate a x = d

c) in the case of x> d

for this case

     all the charge is inside the gaussian surface .  We look for the relationship between volumetric density and surface density

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multiply by the thickness d

       σ = q d / Ad = Q d / V

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we see that the product of the density voluntarily by plate thickness is the surface charge density

        E = σ / 2ε₀  

d) to the direction of the electric field as the charge is positive is leaving the plate

5 0
3 years ago
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