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julia-pushkina [17]
3 years ago
8

If a ball of radius 0.1 m is suspended in water, density = 997 kg/m^3, what is the volume of water displaced and the buoyant for

ce?
Physics
1 answer:
densk [106]3 years ago
7 0

Answer:

Part A

The volume of water displaced is 4.1887902 × 10⁻³ m³

Part B

The buoyant force is approximately 40.93 N

Explanation:

From the question, we have;

The radius of the ball suspended (barely floating) in the water, r = 0.1 m

The density of the water, ρ = 997 kg/m³

Part A

The volume of the ball = The volume of a sphere = (4/3)·π·r³

∴ The volume of the ball = (4/3) × π × 0.1³ = 0.0041887902 m³ = 4.1887902 × 10⁻³ m³

Therefore;

The volume of water displaced, V = The volume of the ball = 4.1887902 × 10⁻³ m³

The volume of water displaced, V = 4.1887902 × 10⁻³ m³

Part B

The buoyant force = The weight of the water displaced = Mass of the water, m × The acceleration due to gravity, g

The buoyant force = m × g

Where;

g ≈ 9.8 m/s²

The mass of the water, m = ρ × V

∴ m = 997 kg/m³ × 4.1887902 × 10⁻³ m³ = 4.17622383 kg

The buoyant force = 4.17622383 kg × 9.8 m/s² ≈ 40.93 N.

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Resultant displacement = 1222.3 m

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mrs_skeptik [129]

Answer:

1. Largest force: C;  smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

F=K\dfrac{ q_{1}q_{2}}{r^{2}}

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write  

F=\dfrac{k}{r^{2}}

1. Net force on each particle

Let's

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Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}}  - \dfrac{k}{(2d)^{2}}  +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

(b) Force on B

\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}}  - \dfrac{k}{d^{2}}  + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

(C) Force on C

\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}}  + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}

(d) Force on D

\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}}  - \dfrac{k}{(2d)^{2}}  - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}

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In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

F_{A} : F_{B} : F_{C} : F_{D}  =  \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\

2. Ratio of largest force to smallest

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Answer: Hello!

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we know that one hour has 60 minutes, then 0.096 hours has:

0.096h*60mins/1h = 5.8 minutes.

then Lewis needs 5.8 minutes in order to cover 16 miles if his speed is 156 miles per hour.

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