Respuesta :
Answer:
The reading of the experiment made in air is 50 g more than the reading of the measurement made in water.
Explanation:
Knowing that the density of lead is [tex]11,3 g/cm^{3}[/tex] and the volume, we can calculate the true weight of the piece of lead:
[tex]weight_{lead}=\rho _{lead}*V_{lead}=11,3 g/cm^{3} *50 cm^{3} = 565 g[/tex]
When the experiment is done in air, we can discard buoyancy force (due to different densities) made by air because it's negligible and the measured weight is approximately the same as the true weight.
When it is done in water, the effect of buoyancy force (force made by the displaced water) is no longer negligible, so we have to take it into account.
Knowing that the density of water is 1 g per cubic centimeter, and that the volume displaced is equal to the piece of lead (because of its much higher density, the piece of lead sinks), we can know that the buoyancy force made by water is 50 g, opposite to the weight of the lead.
[tex]Weight_{measured}=weight_{lead}-weight_{water}=\frac{(565 g *9.8 m/s^{2} -50 g*9.8 m/s^{2})}{9,8m/s^{2} } = 515 g[/tex]
Now that we have the two measurements, we can calculate the difference:
[tex]Difference= |Weight _{in water}- Weight _{in air}|=|515 g-565 g|=50 g[/tex]
The reading of the experiment made in air is 50 g more than the reading of the measurement made in water.
The difference between the measure of weight of a piece of lead in air and water is due to the buoyancy force of water. Difference between the two readings on the scale is 50
What is volume of lead?
Volume of lead is ratio of the weight it to the density of the lead. Thus,
[tex]V_l=\dfrac{W_}{\rho_l}[/tex]
The value of the density for the lead material is 11.3 gram per cm cubed.
Given information-
The volume of the lead is 50 cm cubed.
The experiment was performed two times, once in the air and once in water.
As the density of the lead is 11.3 gram per cm cubed. Thus, the weight of the lead is,
[tex]W=\rho\times V\\W=11.3\times50\\W=565g[/tex]
Hence, the total weight of the piece of lead is 565 grams.
If the air force is negligible then the weight of the piece of lead is 565 grams when the experiment is done in the air with spring scale.
The density of the water is 1 gram per cubic centimetre of water and the buoyancy force applied by the water on the piece of lead with 50 cubic centimetre volume is 50 g.
Thus, the weight of the piece of lead when the experiment is done in the water with spring scale is,
[tex]W_w=565-50\\W_w=515\rm g[/tex]
Thus, the weight of the piece of lead when the experiment is done in the water with spring scale is, 515 grams.
Difference between the two readings on the scale-
[tex]W_d=565-515\\W_d=50\rm g[/tex]
Hence, the difference between the two readings on the scale is 50 grams.
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