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vampirchik [111]
3 years ago
5

There are 640 acres in a square mile. how many square meters are there in 7.80 acres.

Physics
1 answer:
lukranit [14]3 years ago
8 0

Answer:

On 7.8 acres there are 31565.3 square meters

7.8 acres = 31565.3 m²

Explanation:

Conceptual analysis

Acre, square mile and square meter are area units, so their equivalences are in square units.

Known data

1 (mi)² = 640 acres

1 mi = 1609.34 m

Problem development

We know that 1 mile = 1609.34m, then: (1 mile)² = (1609.34m)²=(1609.34)²m²

We apply the known conversion factors to calculate the square meters (X) in 7.8 acres:

X=7.8 acres *\frac{(1 mi)^{2} }{640 acres} * \frac{(1609.34)^{2}m^{2}  }{(1 mi)^{2} }

We eliminate acres and square miles (mi²) because they are in the numerator and denominator, so the answer is in square meters (m²)

X=31565.3 m²

On 7.8 acres there are 31565.3 square meters

7.8 acres = 31565.3 m²

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A mountain lion jumps to a height of 3.25 m when leaving the ground at an angle of 43.2°. What is its initial speed (in m/s) as
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Recall that

{v_f}^2={v_i}^2+2a\Delta y

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At its maximum height, the lion has 0 vertical velocity, so we have

0={v_i}^2-2gy_{\rm max}

where <em>g</em> is the acceleration due to gravity, 9.80 m/s², and we take the starting position of the lion on the ground to be the origin so that \Delta y=y_{\rm max}-0=y_{\rm max}.

Let <em>v</em> denote the initial speed of the jump. Then

v_i=v\sin(43.2^\circ)=\sqrt{2\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(3.25\,\mathrm m)}\implies\boxed{v\approx11.7\dfrac{\rm m}{\rm s}}

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A pelican hunts fish in the water. As the pelican flies above the water’s surface, why must it aim for the fish in a slightly di
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A large, 68.0-kg cubical block of wood with uniform density is floating in a freshwater lake with 20.0% of its volume above the
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Answer:

a) V = 0.085 m^3

b) m = 17 kg

Explanation:

1) Data given

mb = 68 kg (mass for the block)

20% of the block volume is floating

100-20= 80% of the block volume is submerged

2) Notation

mb= mass of the block

Vw= volume submerged

mw = mass water displaced

V= total volume for the block

3) Forces involved (part a)

For this case we have two forces the buoyant force (B), defined as the weight of water displaced acting upward and the weight acting downward (W)

Since we have an equilibrium system we can set the forces equal. By definition the buoyant force is given by :

B = (mass water displaced) g = (mw) g   (1)

The definition of density is :

\rho_w = \frac{m_w}{V_w}

If we solve for mw we got m_w = \rho_w V_w  (2)

Replacing equation (2) into equation (1) we got:

B = \rho_w V_w g (3)

On this case Vw represent the volume of water displaced = 0.8 V

If we replace the values into equation (3) we have

0.8 ρ_w V g = mg  (4)

And solving for V we have

 V =  (mg)/(0.8 ρ_w g )

We cancel the g in the numerator and the denominator we got

V = (m)/(0.8 ρ_w)

V = 68kg /(0.8 x 1000 kg/m^3) = 0.085 m^3

4) Forces involved (part b)

For this case we have bricks above the block, and we want the maximum mass for the bricks without causing  it to sink below the water surface.

We can begin finding the weight of the water displaced when the block is just about to sink (W1)

W1 = ρ_w V g

W1 = 1000 kg/m^3 x 0.085 m^3 x 9.8 m/s^2 = 833 N

After this we can calculate the weight of water displaced before putting the bricks above (W2)

W2 = 0.8 x 833 N = 666.4 N

So the difference between W1 and W2 would represent the weight that can be added with the bricks (W3)

W3 = W1 -W2 = 833-666.4 N = 166.6 N

And finding the mass fro the definition of weight we have

m3 = (166.6 N)/(9.8 m/s^2) = 17 Kg

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