En un recipiente con mercurio (13.6 gr/cm3) se sumerge completamente un cuerpo de 900 gramos de masa. AL medir su peso aparente este marca 2.94 newton en el dinamómetro. Cual es la densidad del cuerpo (kg/m3)

      ρ = 20.4 kg / m³

Explanation:

For this exercise we use Archimedes' principle, the thrust on a submerged body is equal to the weight of the released liquid

           B = ρ g V_liquid           (1)

let's write the equilibrium equation

           F -W + B = 0

           B = W -F

let's calculate

           B = 0.900 9.8 - 2.94

            B = 5.88 N

From equation 1

            V_{liquid} =  

            V_{liquid} =

            V_{liquid} = 4.41176 10⁻⁵ m³

in this case it is indicated that the body is completely submerged

           V_body = V_liquid = V

     the definition of density is

           ρ = m / V

           ρ =

           ρ = 2.04 10¹ kg / m³

           ρ = 20.4 kg / m³

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.

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.

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 . this process does not change the mass number.

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 -decay. in this the mass number remains same.

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

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards