PART ONE
A beaker of mass 1.4 kg containing 2.3 kg of
water rests on a scale. A 3.5 kg block of a
metallic alloy of density 4000 kg/m^3
is suspended from a spring scale and is submerged
in the water of density 1000 kg/m^3
as shown in the figure.
What does the hanging scale read? The
acceleration of gravity is 9.8 m/s^2
Answer in units of N.

PART TWO
What does the lower scale read?
Answer in units of N.

Draw a free body diagram of the block. There are three forces: weight force mg pulling down, buoyancy force ρVg pushing up, and normal force N pushing up.

Sum of forces in the y direction:

∑F = ma

N + ρVg − mg = 0

N = mg − ρVg

N = (3.5 kg) (9.8 m/s²) − (1000 kg/m³) (3.5 kg / 4000 kg/m³) (9.8 m/s²)

N = 25.7 N

The water pushes up on the block with a buoyancy force of ρVg. According to Newton's third law, the block pushes back down on the water with an equal force of ρVg.

The other forces are weight force Mg pulling down, and normal force N pushing up.

Sum of forces in the y-direction:

∑F = ma

N − ρVg − Mg = 0

N = Mg + ρVg

N = (1.4 kg + 2.3 kg) (9.8 m/s²) + (1000 kg/m³) (3.5 kg / 4000 kg/m³) (9.8 m/s²)

N = 44.8 N

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-11-05T15:25:48+01:00

The reading of the hanging scale when the metallic alloy is submerged in the water is 25.73 N.

The reading on the lower scale is 44.84 N.

The given parameters;

mass of the beaker, = 1.4 kg
mass of water, = 2.3 kg
mass of the metallic alloy = 3.5 kg
density of the alloy, = 4000 kg/m³
density of water, = 1000 kg/m³

The volume of water displaced by the alloy is calculated as follows;

[tex]V = \frac{3.5}{4000} = 0.000875 \ m^3[/tex]

The reading of the hanging scale is the net force experienced by the alloy due to the upthrust on it when placed in the water and it is calculated as follows;

[tex]R = mg - \rho Vg\\\\R = (3.5 \times 9.8) - (1000 \times 0.000875 \times 9.8)\\\\R = 25.73 \ N[/tex]

The reading on the lower scale is due to the weight of the water in the beaker and upthrust on the scale.

[tex]R = g(m_1 + m_2) + \rho Vg\\\\R = 9.8(1.4 + 2.3) + (1000 \times 0.000875 \times 9.8)\\\\R = 44.84 \ N[/tex]

Learn more here:https://brainly.com/question/14124424