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tatyana61 [14]
3 years ago
15

A 4.7-kg solid sphere, made of metal whose density is 4000 kg/m3, hangs by a light cord. When the sphere is immersed in water, w

hat is the tension in the cord
Physics
1 answer:
lara31 [8.8K]3 years ago
4 0

Answer:

T = 34.54 N

Explanation:

First we find the buoyant force acting on the sphere, due to displaced water. For that purpose, we need to find the volume of water displaced by the sphere.

Volume of Water Displaced = V = (Mass of Sphere)/(Density of metal)

V = 4.7 kg/(4000 kg/m³)

V = 0.001175 m³

Now, the buoyant force is given as:

F = (Density of Water)(V)(g)

F = (1000 kg/m³)(0.001175 m³)(9.8 m/s²)

F = 11.52 N

Now, we find the weight of the sphere:

W = mg = (4.7 kg)(9.8 m/s²)

W = 46.06 N

Since, both the tension force (T) and buoyant force act in upward direction, while the weight of sphere act in downward direction. Therefore,

W = T + F

T = W - F

T = 46.06 N - 11.52 N

<u>T = 34.54 N</u>

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Debby and Ben took different routes to travel from Point A to Point E. Debby took the route along A, B, C, D, and E. Ben took th
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Ben's average speed was twice Debby's average speed.

Explanation:

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A 7.8-kg solid sphere, made of metal whose density is 2500 kg/m3, is suspended by a cord. When the sphere is immersed in water (
Klio2033 [76]

Answer:

Density of the liquid = 1470.43 kg/m³

Explanation:

Given:

Mass of solid sphere(m) = 6.1 kg

Density of the metal = 2600 kg/m³

Thus volume of the liquid :

Volume of the sphere = 6.1 kg/2600 kg/m³ = 0.002346 m³

The volume of water displaced is equal to the volume of sphere (Archimedes' principle)

Volume displaced = 0.002346 m³

Buoyant force =

Where

is the density of the fluid

g is the acceleration due to gravity

V is the volume displaced

The free body diagram of the sphere is shown in image.

According to image:

Acceleration due to gravity = 9.81 ms⁻²

Tension force = 26 N

Applying in the equation to find the density of the liquid as:

Thus, the density of the liquid = 1470.43 kg/m³

3 0
2 years ago
NEED HELP WITH THIS PLEASE
Jobisdone [24]

Answer:

A

Explanation:

5 0
3 years ago
In measuring the width of a hair sample, a light of wavelength 500 nm is used. The hair sample is 40 um in radius. With the scre
sergejj [24]

Answer:

The distance of the second dark band away from the central bright spot be located is 5.625\times10^{-2}\ m

Explanation:

Given that,

Wave length = 500 nm

Radius d= 40\ \mu m

Distance from the hair sample D= 6 m

We need to calculate the distance of the second dark band away from the central bright spot be located

\sin\theta=\dfrac{y}{D}

\sin\theta=\dfrac{y}{6}

Using formula for dark fringe

(n-\dfrac{1}{2})\lambda=2d\sin\theta

Put the value into the formula

(2-\dfrac{1}{2})\times500\times10^{-9}=2\times40\times10^{-6}\times\dfrac{y}{6}

y=\dfrac{(2-\dfrac{1}{2})\times500\times10^{-9}\times6}{2\times40\times10^{-6}}

y=0.05625\ m

y=5.625\times10^{-2}\ m

Hence, The distance of the second dark band away from the central bright spot be located is 5.625\times10^{-2}\ m

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3 years ago
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