Physics, 24.12.2019 21:31 soccer816

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 45.0 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?

Answers: 1

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Answers

Answer from: cynone8254

(a). The mass of displaced water is $41.3\times10^{-3}\ kg$

(b). The volume of the bone is $41.3\times10^{-6}\ m^3$

(c). The average density is 1089.58 kg/m³.

Explanation:

Given that,

Mass of bone in air = 45.0 g

Mass of bone in water = 3.60 g

(a). We need to calculate the mass of displaced water

Using buoyant force

The buoyant force equal to the difference in apparent weight in air and water

$F_{b}=m_{dw}g$

$m_{dw}g=m_{a}g-m_{w}g$

Put the value into the formula

$m_{dw}g=45\times10^{-3}\times9.8-3.60\times10^{-3}\times9.8$

$m_{dw}g=0.405$

$m_{dw}=\dfrac{0.405}{9.8}$

$m_{dw}=0.0413\ kg$

$m_{dw}=41.3\times10^{-3}\ kg$

$m_{dw}=41.3\ g$

(b). We need to calculate the volume of the bone

Using formula of volume

$V_{dw}=\dfrac{m_{dw}}{\rho_{w}}$

We know that,

$V_{dw}=V_{b}$

Put the value into the formula

$V_{dw}=\dfrac{41.3\times10^{-3}}{1000}$

$V_{dw}=41.3\times10^{-6}\ m^3$

(c). We need to calculate the average density

Using formula of density

$\rho_{b}=\dfrac{m_{a}}{V_{b}}$

Put the value into the formula

$\rho_{b}=\dfrac{45.0\times10^{-3}}{41.3\times10^{-6}}$

$\rho_{b}=1089.58\ kg/m^3$

Hence, (a). The mass of displaced water is $41.3\times10^{-3}\ kg$

(b). The volume of the bone is $41.3\times10^{-6}\ m^3$

(c). The average density is 1089.58 kg/m³.

Answer from: Quest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.

$_a^z\textrm{x}\rightarrow _{z-1}^a\textrm{y}+_{+1}^0e$

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

Answer from: Quest

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.