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lina2011 [118]
2 years ago
13

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 (

of density 1000 kg/m3), what is the tension in the cord
Physics
1 answer:
Klio2033 [76]2 years ago
3 0

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

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A 2 kg watermelon is dropped from a 4 m tall roof. How fast will the watermelon be moving as it strikes the ground below?
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(see the attached graphic)

We have

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\vec v_{D/W}=\left(7.0\dfrac{\rm m}{\rm s}\right)(\cos\theta\,\vec\imath+\sin\theta\,\vec\jmath)

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\vec v_{W/E}=\left(5.0\dfrac{\rm m}{\rm s}\right)(\cos0^\circ\,\vec\imath+\sin0^\circ\,\vec\jmath)

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Then by setting components equal, we have

\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta+5.0\dfrac{\rm m}{\rm s}=0

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We only care about the direction for this question, which we get from the first equation:

\left(7.0\dfrac{\rm m}{\rm s}\right)\cos\theta=-5.0\dfrac{\rm m}{\rm s}

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\theta=\cos^{-1}\left(-\dfrac57\right)\text{ OR }\theta=360^\circ-\cos^{-1}\left(-\dfrac57\right)

or approximately 136º or 224º.

Only one of these directions must be correct. Choosing between them is a matter of picking the one that satisfies <em>both</em> equations. We want

\left(7.0\dfrac{\rm m}{\rm s}\right)\sin\theta=-v

which means <em>θ</em> must be between 180º and 360º (since angles in this range have negative sine).

So the ducks must fly (relative to the air) in a direction 224º relative to the positive horizontal direction, or about 44º South of West.

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