A beaker weighs 0.4N when empty and1.4N when filled with water what does ot weigh when filled with brine of density 1.2 g/cm3

1.6 N

Explanation:

The weight of an object is calculated as:

W = g*m

where:

g = gravitational acceleration = 9.8m/s^2

m = mass of the object.

We also know that:

mass = density*volume.

or:

m = d*v

Let's start:

The weight of the beaker is 0.4N

And the weight of the beaker filled with water is 1.4N

Then the weight of the water alone will be:

1.4N - 0.4N = 1N = (d*v)*9.8m/s^2

And we know that the density of the water is:

1 g/cm^3

But we are working with Newtons, then we need to rewrite this with kilograms as the mass unit, we can use that:

1000g = 1kg

Now we can rewrite the density as:

d = 1 g/cm^3 = 1*(1/1000) kg/cm^3 = 0.001 kg/cm^3

Replacing that in the above equation, we get:

With this, we can find the volume that the water occupies.

W = 1 N = v* 0.001 kg/cm^3*9.8m/s^2

       1 N /( 0.001 kg/cm^3*9.8m/s^2 ) = 102.04 cm^3.

Now, when we fill it with a brine with a density of 1.2 g/cm^3, the mass of this brine in a volume of  102.04 cm^3 be:

M = (1.2 g/cm^3)*( 102.04 cm^3) = 122.448 g

Rewriting this in kg we get:

M = 122.448 g = (122.448/1000) kg = 0.122448 kg

Then the weight of this brine is:

M = 0.122448 kg*9.8m/s^2 = 1.2 N

And the beaker weighs 0.4N, then the beaker filled with this brine will weight:

1.2 N + 0.4N = 1.6 N

im guessing the west of station b