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Viktor [21]
3 years ago
10

A valuable statuette from a Greek shipwreck lies at the bottom of the Mediterranean Sea. The statuette has a mass of 10,566 g an

d a volume of 4,064 cm3. The density of seawater is 1.03 g/mL.
a. What is the weight of the statuette?
b. What is the mass of displaced water?
c. What is the weight of displaced water?
d. What is the buoyant force on the statuette?
e. What is the net force on the statuette?
f. How much force would be required to lift the statuette?
Physics
1 answer:
leonid [27]3 years ago
4 0

Answer:

A) W = 103.55 N

B) mass of displaced water = 4186 g

C) W_displaced water = 41.06 N

D) Buoyant force = 41.06 N.

E) ZERO

F) 62.54 N

Explanation:

We are given;

mass of statuette;m = 10,566 g = 10.566 kg

volume = 4,064 cm³

Density of seawater;ρ = 1.03 g/mL = 1.03 g/cm³

A) The dry weight of the statuette can be calculated as;

W = mg

So;

W = 10.556 × 9.81

W = 103.55 N

B) Mass of displaced water is calculated from;

Density = mass/volume

So, mass = Density × Volume

m = 1.03 × 4,064 = 4186 g

C) Weight of displaced water is given by;

W_displaced water = (m_displaced water) × g

W_displaced water = 4.186 kg × 9.81 m/s^2 = 41.06 N

D) The buoyant force is the same as the weight of the displaced water.

Thus, Buoyant force = 41.06 N.

E) The apparent weight of the statuette is calculated from;

Apparent weight = Dry weight - Weight of displaced water

Apparent weight = 103.6 N - 41.06 N = 62.54 N. It is sitting on the bottom of the sea, so the sea floor is providing an opposite force that is equal but opposite the weight so that the net force on the statuette is zero. Since It has zero acceleration, in any direction, hence the net force on it is zero.

F. From E above, The Force required to lift the statuette = 62.54 N

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GarryVolchara [31]

Answer:

1) 0.1584 m

2) To allow for expansion without derailment

3) 0.101376 m

4) 213.675 °C

5) 266.67 m

6) 8.33 × 10⁻⁶ /°C

7) The alloy meets the requirement

8) 1.95 × 10⁻³ /°C

9) 32.095 m

10) -12157.72°C

Explanation:

1) Equation for the coefficient of linear expansion = \frac{\Delta L}{L} = \alpha _L \Delta T

Where:

ΔL = Change in length = Required

L = Initial length = 1.32 × 10⁴ m

\alpha _L = Coefficient of linear expansion of steel = 12 × 10⁻⁶ /°C

ΔT = Change in temperature = 37°C - 27°C = 10°C

Plugging the values in the equation for the temperature expansion of steel, we have m;

ΔL = L × \alpha _L ×ΔT = 1.32 × 10⁴ × 12 × 10⁻⁶ × 10  = 0.1584 m

2. Here we have that by segmenting railroad tracks into short pieces, the expansion of the metal tracks with temperature can be absorbed by the gaps between the segment without distorting the shape and direction (pattern) of the tracks

3. Here we have;

\alpha _L = Coefficient of linear expansion of iron = 12 × 10⁻⁶ /°C

ΔT = Temperature change = 27°C - 3°C = 24°C

L = Height of the Eiffel Tower = 352 meters

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Therefore, the height of the Eiffel Tower changes from 352 m to about 352.101376 m each year, with an average change in height experienced each year = 0.101376 m

4. Here, we have

L = 13.0 ft

ΔL = 1 in.

\alpha _L = 30 × 10⁻⁶ /°C

ΔT = Required temperature change

From  \frac{\Delta L}{L} = \alpha _L \Delta T

\Delta T =\frac{\Delta L}{L \times  \alpha _L} = \frac{1}{156 \times 30 \times 10^{-6}} = 213.675^{\circ}C

5. Here, we have;

L = \frac{\Delta L}{\alpha _L \Delta T}

∴ L = 1/(150×25 × 10⁻⁶) = 266.67 m

The bars original length = 266.67 m

6. Here we have;

\alpha _L = \frac{\Delta L}{L \times \Delta T}

Where:

ΔL = 3.00 - 3.002 = 0.002 m

L = 3.00 m

ΔT = 110°C - 30°C = 80°C

∴ \alpha _L = 0.002/(3.00 × 80) = 8.33 × 10⁻⁶ /°C

7. Here we have;

ΔL = L × \alpha _L ×ΔT = 3 × 8.33 × 10⁻⁶ × 210 = 0.00525 m

Therefore, final length = 3.00 m + 0.00525 m = 3.00525 m

Since 3.00525 m < 3.017 m hence the alloy meets the requirement.

8. Here, we have

L = 3.2 m

ΔL = 0.5 m

ΔT = 84°C - 24°C = 60°C

∴ \alpha _L = 0.5/(3.2 × 60) = 1.95 × 10⁻³ /°C

The coefficient of linear expansion of the material from which the rod is made = 1.95 × 10⁻³ /°C

9. Here, we have

Length of steel girder, L = 32.10 m

ΔT = 8°C - 22°C = -14°C

\alpha _L = 12 × 10⁻⁶ /°C

ΔL = L × \alpha _L ×ΔT

Hence ΔL = 32.1 × 12 × 10⁻⁶× -14 = -0.0054 m

New length = 32.1 - 0.0054 = 32.095 m

10. Here we have;

ΔL = 92.6 cm - 123 cm  = -30.4 cm

\alpha _L = 2.0 × 10⁻⁵ /°C

L = 123 cm

∴ \Delta T =\frac{\Delta L}{L \times  \alpha _L} = \frac{-30.4}{123 \times 2.0 \times 10^{-5}} = -12357.724^{\circ}C

Therefore, the temperature will be 200 - 12357.724 = -12157.72°C.

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A marble dropped in still water will create circular ripples or waves. the radius of each circular wave will increase 4 centimet
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The circumference of the circle after t seconds = 25.12 t

Given data,

The radius of each circular wave will increase by four centimeters per second (cm/s)

In 1 second of circular wave radius = 4cm

In 1*t second of circular wave radius = 4t cm

 

As a result, the radius of the circular wave after t seconds is 4t cm.

We already know that the circumference of a circle is given by = 2πr, where r is the circle's radius.

As a result, the radius of a circular wave after t seconds is = 2π * radius of a circular wave after t seconds = 2π * 4t

                                                    = 8πt

( Taking π = 3.14 )

As a result, the circumference of the circle after t seconds = 8πt

                                                                                                     = 3.14t

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Find more on circumference at : brainly.in/question/49774764

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5 0
2 years ago
Why are weight and mass used synonymously on earth?
Lisa [10]
Weight and mass are used synonymously on earth because the value of g is constant on the earth because the weight of a body is the amount of force exerted by the Earths gravity on an object of finite mass. And, the mass of an object gives the amount of matter in the body and is measured in kilograms. 
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3 years ago
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vaieri [72.5K]

Answer:

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Explanation:

Given data

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To

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Solution

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Answer:

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