Physics metal and air density problem

[mathelord]

a metal of density x weighs y in air,when immersed in a liquid of density z,what is it's apparent weight.please i need explanations about this

 

[da_willem]

The weight, other than the mass, of an object is the force gravity excerts on it. So although your mass remains the same, your weight would be different on e.g. the moon. You could also include the buoyancy force in the weight so when an object is immersed in a fluid e.g. it's weight decreases, as you might notice when you go swimming

So you have to find the net force on an object immersed in a liquid. Do you know the expression for the buoyancy force, Archimedes law?

 

[thepatientmental]

The apparent weight of the metal when immersed in the liquid can be calculated using the formula:

Apparent weight = Weight in air - Buoyant force

The weight in air is given as y, and the buoyant force can be calculated using the formula:

Buoyant force = Volume of the object x Density of the liquid x Acceleration due to gravity

The volume of the object can be calculated using its density (x) and its weight in air (y):

Volume = Weight in air / Density

Substituting these values into the buoyant force formula, we get:

Buoyant force = (Weight in air / Density) x Density of the liquid x Acceleration due to gravity

= (y / x) x z x g

where g is the acceleration due to gravity.

Now, we can plug in this value for buoyant force into the formula for apparent weight:

Apparent weight = Weight in air - Buoyant force

= y - (y / x) x z x g

= y (1 - z / x) x g

Therefore, the apparent weight of the metal when immersed in the liquid is y (1 - z / x) x g. This means that the apparent weight will be less than the weight in air, since the buoyant force acts in the opposite direction of gravity. The greater the difference between the densities of the metal and the liquid, the greater the reduction in apparent weight will be. I hope this explanation helps to clarify the concept of apparent weight in this physics problem.