Question

A block of density pb = 9.50 times 10^2 kg/m^3 floats face down in a fluid of density pt = 1.30 times 10^3 kg/m^3. The block has height H = 8.00 cm. By what depth ft is the block submerged? If the block is held fully submerged and then released, what is the magnitude of its acceleration?

Answer 1

The depth and acceleration are 0.1919291 ft and 3.61 m/s².

Step-by-step explanation:

Given that,

Density of block
`\rho_(b) =9.50*10^2\ kg/m^3`

Density of fluid
`\rho_(t) =1.30*10^3\ kg/m^3`

We need to calculate the depth

Using balance equation

`mg=\rho g V`....(I)

We know that,

The density is

`\rho=(m)/(V)`

`m=\rh0* V`

Put the value of m in equation (I)

`\rho_(b)* V* g=\rho_(t)* g* V`

`\rho_(b)* A* H* g=\rho_(t)* g* A* h`

`h=(\rho_(b)H)/(\rho_(t))`

Put the value into the formula

`h=(9.50*10^2*8.00*10^(-2))/(1.30*10^3)`

`h= 5.85\ cm`

`h=0.1919291\ ft`

We need to calculate the acceleration

Using formula of net force

`F_(net)=\rho_(t)* g* V- \rho_(b)* g* V`

`ma=\rho_(t)* g* V- \rho_(b)* g* V`

`\rho_(b)* V* a=\rho_(t)* g* V- \rho_(b)* g* V`

`a=((\rho_(t))/(\rho_(b))-1)g`....(II)

Put the value in the equation (II)

`a=((1.30*10^3)/(9.50*10^2)-1)*9.8`

`a=3.61\ m/s^2`

Hence, The depth and acceleration are 0.1919291 ft and 3.61 m/s².