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musickatia [10]

3 years ago

10

You are visiting your friend Fabio's house. You find that, as a joke, he filled his swimming pool with Kool-Aid, which dissolved

perfectly into the water. However, now that you want to swim, you must remove all of the Kool-Aid contaminated water. The swimming pool is round, with a 9 foot radius. It is 8 feet tall and has 7.5 feet of water in it. How much work is required to remove all of the water by pumping it over the side? Use the physical definition of work, and the fact that the density of the Kool-Aid contaminated water is σ=63.5lbs/ft3

2 answers:

oksian1 [2.3K]3 years ago

3 0

Answer:

Explanation:

The volume of contaminated water

= cross sectional area x height of water level

3.14 x 9 x 9 x 7.5 ft³

= 1907.55 ft³

mass = density x volume

= 1907.55 x 63.5 lbs

m = 121129.425 lbs

This mass has to be raised to the height of 8 ft before evacuation .

There is a rise of centre of mass of

8 - 7.5/2 ft

h = 4.25 ft

Energy required

= mgh

= 121129.425 x 32 x 4.25

= 16473601.8 unit.

skelet666 [1.2K]3 years ago

3 0

Answer:

Workdone= 8.08×10^3Joules

Explanation:

dV= pir^2 dx

dV = pi×9^2 = 81pi

dF= density× pi× r^2 dx

dF= 63.5 × 81pi dx= 5143.5pi dx

W= Integral from7.5ft of water in the pool(8 -×)dx

W= 5143.5pi(8-7.5)

W= 5143.5 × 3.142 × 0.5

W= 8080.44Joules = 8.08×10^3Joules

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A manufacturer claims that a carpet will not generate more than 5.8 kV of static electricity What magnitude of charge would have

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4.4×10⁻⁷ Coulomb

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V = Voltage = 5.8 kV

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$\frac{Q}{V}=\frac{A\epsilon_0}{d}\\\Rightarrow \Q=V\frac{A\epsilon_0}{d}\\\Rightarrow Q=5.8\times 10^3\frac{0.024\times 8.85\times 10^{-12}}{0.0028}\\\Rightarrow Q=4.4\times 10^{-7}\ C$

Magnitude of charge transferred between a carpet and a shoe is 4.4×10⁻⁷ Coulomb.

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3 years ago

A man takes 20 seconds to climb 5m up a ladder. He weighs 720N. Calculate the power he must deliver to do this.

Inessa05 [86]

Answer:

180 W

Explanation:

The work done by the man against gravity is equal to its gain in gravitational potential energy:

$W=mg\Delta h$

where

(mg) = 720 N is the weight of the man

$\Delta h= 5 m$ is the change in height

Substituting,

$W=(720)(5)=3600 J$

The power he must deliver is given by

$P=\frac{W}{t}$

where

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t = 20 s is the time taken

Substituting,

$P=\frac{3600}{20}=180 W$

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3 years ago

Suppose you first walk 12.0 m in a direction 20? west of north and then 20.0 m in a direction 40.0? south of west. how far are y

Gnesinka [82]

The representation of this problem is shown in Figure 1. So our goal is to find the vector $\overrightarrow{R}$ . From the figure we know that:

$\left | \overrightarrow{A} \right |=12m \\ \\ \left | \overrightarrow{B} \right |=20m \\ \\ \theta_{A}=20^{\circ} \\ \\ \theta_{B}=40^{\circ}$

From geometry, we know that:

$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$

Then using vector decomposition into components:

$For \ A: \\ \\ A_x=-\left | \overrightarrow{A} \right |sin\theta_A=-12sin(20^{\circ})=-4.10 \\ \\ A_y=\left | \overrightarrow{A} \right |cos\theta_A=12cos(20^{\circ})=11.27 \\ \\ \\ For \ B: \\ \\ B_x=-\left | \overrightarrow{B} \right |cos\theta_B=-20cos(40^{\circ})=-15.32 \\ \\ B_y=-\left | \overrightarrow{B} \right |sin\theta_B=-20sin(40^{\circ})=-12.85$

Therefore:

$R_x=A_x+B_x=-4.10-15.32=-19.42m \\ \\ R_y=A_y+B_y=11.27-12.85=-1.58m$

So if you want to find out <span>how far are you from your starting point you need to know the magnitude of the vector $\overrightarrow{R}$ , that is:
</span>
$\left | \overrightarrow{R} \right |=
\sqrt{R_x^2+R_y^2}=\sqrt{(-19.42)^2+(-1.58)^2}=\boxed{19.48m}$

Finally, let's find the <span>compass direction of a line connecting your starting point to your final position. What we are looking for here is an angle that is shown in Figure 2 which is an angle defined with respect to the positive x-axis. Therefore:

</span> $\theta_R=180^{\circ}+tan^{-1}(\frac{\left | R_y \right |}{\left | R_x \right |}) \\ \\ \theta_R=180^{\circ}+tan^{-1}(\frac{1.58}{19.42}) \\ \\ \theta_R=180^{\circ}+4.65^{\circ}=185.85^{\circ}$

6 0

3 years ago

3. A 2kg wooden block whose initial speed is 3 m/s slides on a smooth floor for 2 meters before it comes to a

serious [3.7K]

Answer:

Calculating Coefficient of friction is 0.229.

Force is 4.5 N that keep the block moving at a constant speed.

Explanation:

We know that speed expression is as $\mathrm{V}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 . \mathrm{a} . \Delta \mathrm{s}$ .

Where, ${V}_{i}$ is initial speed, V is final speed, ∆s displacement and a acceleration.

Given that,

${V}_{i}$ =3 m/s, V = 0 m/s, and ∆s = 2 m

Substitute the values in the above formula,

$0=3^{2}-2 \times 2 \times a$

0 = 9 - 4a

4a = 9

$a=2.25 \mathrm{m} / \mathrm{s}^{2}$

$a=2.25 \mathrm{m} / \mathrm{s}^{2}$ is the acceleration.

Calculating Coefficient of friction:

$\mathrm{F}=\mathrm{m} \times \mathrm{a}$

$\mathrm{F}=\mu \times \mathrm{m} \times \mathrm{g}$

Compare the above equation

$\mu \times m \times g=m \times a$

Cancel "m" common term in both L.H.S and R.H.S

$\text { Equation becomes, } \mu \times g=a$

$\text { Coefficient of friction } \mu=\frac{a}{g}$

$\mathrm{g} \text { on earth surface }=9.8 \mathrm{m} / \mathrm{s}^{2}$

$\mu=\frac{2.25}{9.8}$

$\mu=0.229$

Hence coefficient of friction is 0.229.

calculating force:

$\text { We know that } \mathrm{F}=\mathrm{m} \times \mathrm{a}$

$\mathrm{F}=2 \times 2.25 \quad(\mathrm{m}=2 \mathrm{kg} \text { given })$

F = 4.5 N

Therefore, the force would be <u>4.5 N</u> to keep the block moving at a constant speed across the floor.

7 0

2 years ago