Respuesta :
Answer:
The density of the object is 8000 kg/m^3
Explanation:
Weight in air = 7.84 n
Weight in water = 6.86 N
density of water = 1000 kg/m^3
Let d be the density of the object
According to the Archimedes principle, when a body is immersed in a liquid partly or wholly, it experiences an upward force which is called buoyant force. The buoyant force is equal to the loss in weight of the body.
Loss in weight of the object = Weight of object in the air - the weight of an object in the water
Loss in weight = 7.84 - 6.86 = 0.98 N
The volume of body x density of water x g = 0.98
Let V be the volume of the body
V x 1000 x 9.8 = 0.98
V = 10^-4 m^3
Weight in air = Volume of body x density of body x g
7.84 = 10^-4 x d x 9.8
d = 8000 kg/m^3
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#4208.
The density of the object is ([tex]\rho_{ob}[/tex])= 80,000 [tex]kg/m^3[/tex]
How can we calculate the value of the density of the object?
To calculate the density, we have to calculate the buoyancy force that the water exerts on the object. To calculate the buoyancy force we use the formula,
[tex]B= W_{air}- W_{water}[/tex]
Here we are given,
[tex]W_{air}[/tex] = The weight of the object in air = 7.84 N
[tex]W_{water}[/tex]= The weight of the object in water= 6.86 N
We have to calculate the buoyancy force =B
Now we put the known values in the above equation, we get
[tex]B= 7.84-6.86[/tex] = 0.98 N.
We know that the buoyant force depends on the volume of water displaced by the Volume of the object, so the formula stands,
[tex]B= \rho_{water} \times V_{ob}\times g[/tex]
Or, [tex]V_{ob}=\frac{B}{g\times \rho_{water}}[/tex]
Here we are given,
B= The buoyancy force = 0.98N.
[tex]\rho_{water}[/tex]= The density of water = 1000 [tex]Kg/m^3[/tex]
g= gravitational acceleration= 9.81[tex]m/s^2[/tex]
We have to calculate the volume of the object= [tex]V_{ob}[/tex]
Now we put the known values in the above equation, we get
[tex]V_{ob}=\frac{B}{g\times \rho_{water}}[/tex]=[tex]\frac{0.98}{9.81\times 1000}[/tex]=[tex]1.0\times 10^{-5} m^3[/tex].
So, now the volume of the object ([tex]V_{ob}[/tex])= [tex]1.0\times 10^{-5} m^3[/tex]
The weight of the object in the air can be calculated using the following formula, [tex]W_{ob}= m_{ob} \times g[/tex]
[tex]m_{ob}[/tex]=the mass of the object.
From this equation we can calculate the mass of the object,
[tex]m_{ob}=\frac{W_{ob}}{g}[/tex]= [tex]\frac{7.84}{9.81}[/tex]=0.80Kg.
(We know that the weight of the object in the air is the original weight of the object, so [tex]W_{ob} = W_{air}[/tex]=7.84N )
So, the mass of the object is ([tex]m_{ob}[/tex])= 0.80 Kg.
Now, let us consider the density of the object is [tex]\rho_{ob}[/tex]
From the definition of the density we can simply know that,
[tex]\rho_{ob}=\frac{m_{ob}}{ V_{ob}}[/tex]
Or,[tex]\rho_{ob}=\frac{0.80 }{1.0\times 10^{-5} }[/tex]
Or.[tex]\rho_{ob}=8\times 10^{4} Kg/m^3[/tex]
Or, [tex]\rho_{ob}=80,000 Kg/m^3[/tex]
From the above calculation we can define that the density of the object is 80,000 [tex]kg/m^3[/tex]
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