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Murrr4er [49]
3 years ago
12

Bowling balls are roughly the same size, but come in a variety of weights. Given its official radius of roughly 0.110 m, calcula

te the heaviest bowling ball that will float in a fluid of density 1.100
Physics
1 answer:
svetlana [45]3 years ago
7 0

To solve this problem we will apply the concepts related to the balance of forces. In this case the force caused by the weight and the buoyancy force on the Fluid. Both are forces that start from Newton's second law and can be expressed as

Weight

F_W = mg

Here,

m = mass

g = Gravity

Buoyant Force

F_B = \rho_{fluid}V_{pin}g

Here

\rho_{fluid} = Density of Fluid

V_{pin}=Volume

g = Gravity

By equilibrium we have that

F_W = F_B

mg=\rho_{fluid}V_{pin}g

m=\rho_{fluid}V_{pin}

m=\rho_{fluid}(\frac{4}{3} \pi r^3 )

Replacing,

m=(1.1*10^3) [(4/3)\pi(0.11)^3]

m =6.132 Kg

Now the Weight of the balls would be

F = mg

F = 6.132 (9.8)

F = 60.0936N

Therefore the  heaviest bowling ball that will float in this fluid is 60.0936N

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Thomas and John are carrying a 43kg cylinder head on a 510cm X 510mm board. The cylinder head with dimensions of 43cm X 250mm li
tatyana61 [14]

John carry the heaviest load.

<h3>How to find out who is carrying the heavy load?</h3>

Write down given data from questions:

Board=510cm X 510mm.

Cylinder head with dimensions=43cm X 250mm.

Cylinder lies across the board 210cm from john.

Find out: Who is carry the heaviest load?​

Calculation:

We assume that mass of cylinder head = x kg

Then weight=x x 9*81

                 W=9.81x  Newton.

Weight per unit length= Weight/Total leanth

Weight per unit length= 9.81x/43

(w/l)=0.23x N/cm

From equation contition: (F_{J} +F_{T} =9.81x n)

F_{T} (510)=9.81x  (210+21.5)

F_{T} (510)=9.81 x (231.5)

F_{T} =4.452x N.

F_{J} =9.81x-4.452x

F_{J} =5.358 xN

Therefore F_{J} \geq F_{T}

To learn more about mass per unit length, refer to:

brainly.com/question/24180692

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2 years ago
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Answer:

Explanation:

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T₀ is time elapsed in moving reference , T time elapsed in stationary reference.

Here T₀ = 1 second

T = 1/√ 1-0.9² = 1/.4358 = 2.3 second

So 2.3 second will pass for each second on moving reference.

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