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3 years ago

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An aluminum block has a mass of 0.25 kg and a density of 2700 kg/m3. a) Calculate the volume of the aluminium block. b) Determin

e the buoyant force exerted on this block when it is completely submerged in water of density 1000 kg/m3

1 answer:

Hatshy [7]3 years ago

7 0

Answer:

a

The volume of the aluminum is $V = 9.25*10^{-5}m^3}$

b

The buoyant force exerted is $F_B =0.9N$

From the question we are told that

The mass of the aluminum is $M_A = 0.25Kg$

The density of the aluminum is $\rho_A = 2700 kg/m^3$

Generally volume is mathematically represented as

$V = \frac{M_A}{\rho_A}$

Substituting values accordingly

$V =\frac{0.25}{2700} =9.25*10^{-5}m^3$

Generally buoyant force is mathematically represented as

$F_B = V \rho_wg$

Where V is the volume of the aluminum $=9.25*10^{-5}m^3$

$\rho_w$ is the density of water $=1000kg/m^3$

g is the acceleration due to gravity = $=9.8m/s^2$

Substituting these values accordingly we have

$F_B = 9.25*10^{-5} * 1000 * 9.8$

$=0.9N$

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Read 2 more answers

A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi

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To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

$W = F * d$

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

$d=h*N$ , for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

$V_f^2-V_i^2 = 2a\Delta X$

Where,

$V_f =$ Final velocity

$V_i =$ Initial Velocity

a = Acceleration

$\Delta X =$ Displacement

PART A) For the particular case of work we know then that,

$W = F*d$

$W = m*g*(h*N)$

$W = 50*9.8*(0.3*30)$

$W = 4.41kJ$

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

$V_f^2-V_i^2 = 2a\Delta X$

Here,

$V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow$ 3 steps in one second

$v_f = 0$

Replacing,

$V_f^2-V_i^2 = 2a\Delta X$

$0-0.9^2=2a(30*0.3)$

Re-arrange for a,

$a = -\frac{0.9^2}{2*30*0.3}$

$a = -45*10^{-3}m/s^2$

At this point we can calculate the time, which is,

$t = \frac{\Delta V}{a}$

$t = \frac{0-0.9}{-45*10^{-3}}$

$t = 20s$

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

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3 years ago