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CaHeK987 [17]
4 years ago
15

An iron anchor of density 7810.00 kg/m^3 appears 252 N lighter in water than in air. (a) What is the volume of the anchor?(b) Ho

w much does it weigh in air?
Physics
1 answer:
RSB [31]4 years ago
8 0

Answer:

(a)The volume of the iron anchor = 25.2 m³

<em>(b)The mass of iron = 196812 kg</em>

Explanation:

<em>Density:</em> This is the ratio of the mass of a body to its volume. The S.I unit of density is <em>kg/m³</em>

(a)

From Archimedes principle, the iron will displace a volume of water that is equal to its volume. And the weight of water displaced is equal to loss in weight or upthrust.

Therefore, Upthrust = lost in weight = weight of water displace.

Therefore, Volume of the iron = mass of water displace/density of water

Vi = m₁/D₁........................... Equation 1

Where Vi = volume of iron, m₁ = mass of water displaced, D₁ = Density of water.

<em>Given: upthrust = 252 N, ∴ m₁ = 252/10 = 25.2 kg.</em>

<em>Constant: D₁ = 1.0 kg/m³</em>

<em>Substituting these values into equation 1,</em>

Vi = 25.2/1.0

Vi = 25.2 m³

Therefore the volume of the iron anchor = 25.2 m³

(b)

<em>Density of the iron = mass of the iron/volume of the iron.</em>

<em>Therefore, Mass of the iron = Density of the iron × volume of the iron...........................................equation 2</em>

<em>Given: Density of the iron = 7810 kg/m³, Volume of the iron =  25.2 m³</em>

<em>Substituting these values into equation 2,</em>

<em>Mass of the iron = 7810×25.2</em>

<em>Mass of the iron  = 196812 kg.</em>

<em>Therefore the mass of iron = 196812 kg</em>

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Answer:

(472i + 80.3j) m

Explanation:

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The final displacement of the golf ball from the tee:

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r_final = (300 + 172)i m + 80.3j m

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3 years ago
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kumpel [21]

Answer:

72.53 mi/hr

Explanation:

From the question given above, the following data were obtained:

Vertical distance i.e Height (h) = 8.26 m

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Horizontal velocity (u) =?

Next, we shall determine the time taken for the car to get to the ground.

This can be obtained as follow:

Height (h) = 8.26 m

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Time (t) =?

h = ½gt²

8.26 = ½ × 9.8 × t²

8.26 = 4.9 × t²

Divide both side by 4.9

t² = 8.26 / 4.9

Take the square root of both side by

t = √(8.26 / 4.9)

t = 1.3 s

Next, we shall determine the horizontal velocity of the car. This can be obtained as follow:

Horizontal distance (s) = 42.1 m

Time (t) = 1.3 s

Horizontal velocity (u) =?

s = ut

42.1 = u × 1.3

Divide both side by 1.3

u = 42.1 / 1.3

u = 32.38 m/s

Finally, we shall convert 32.38 m/s to miles per hour (mi/hr). This can be obtained as follow:

1 m/s = 2.24 mi/hr

Therefore,

32.38 m/s = 32.38 m/s × 2.24 mi/hr / 1 m/s

32.38 m/s = 72.53 mi/hr

Thus, the car was moving at a speed of

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3 years ago
Bart stole a watermelon in ran 5,000 feet from the cops to chase lasted 0.1 hours how fast was Bart running in miles per hour...
valentinak56 [21]
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3 0
3 years ago
calculate the diameter of a silver wire of length 75cm , which is extended by 1.85mm when a 10kg mass is suspended from it's end
sdas [7]

Answer:0.8\ mm

Explanation:

Given

length of wire l=75\ cm

change in length \Delta l=1.85\ mm

mass of wire m=10\ kg

Young's modulus for silver E=7.9\times 10^{10}\ N/m^2

load on wire F=mg

F=10\times 9.8=98\ kg

change in length is given by

\Delta l=\dfrac{Pl}{AE}

Where A=area of cross-section

A=\dfrac{Pl}{\Delta lE}

A=\dfrac{98\times 0.75}{1.85\times 10^{-3}\times 7.9\times 10^{10}}

A=\dfrac{73.5}{14.615\times 10^{7}}

A=5.029\times 10^{-7}\ m^2

also wire is the shape of cylinder so cross-section is given by

A=\dfrac{\pi d^2}{4}=5.029\times 10^{-7}\ m^2

\Rightarrow d^2=\dfrac{5.029\times 10^{-7}\times 4}{\pi }

\Rightarrow d^2=64.02\times 10^{-8}

\Rightarrow d=8\times 10^{-4}\ m

\Rightarrow d=0.8\ mm

4 0
3 years ago
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