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Serggg [28]
3 years ago
8

A Submerged Ball Part A A ball of mass mb and volume V is lowered on a string into a fluid of density Pi (Figure 1) Assume that

the object would sink to the bottom if it were not supported by the string. What is the tension T in the string when the ball is fully submerged but not touching the bottom as shown in the figure? Express your answer in terms of any or all of the given quantities and g, the magnitude of the acceleration due to gravity
Note:
a ball is suspended in a fluid
Physics
1 answer:
frozen [14]3 years ago
5 0

Answer:

T = mg -   \rho_f×V×g

Explanation:

Here we have the mass of the ball is given as m_b

The density of the fluid = \rho_f

The volume of the ball = V

Acceleration due to gravity = g

According to Archimedes's principle, the upthrust = Weight of fluid displaced

Tension in string = T = weight of ball - upthrust

T = mg -   \rho_f×V×g

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The volume of a warmed part of the air is reduced and its density increases.

Explanation:

In a convective form of heat transfer, the volume of a warmed part of air is not reduced and its density does not increase.

During convection, heat causes the warm part of the air to expand and its volume increases. When volume increases, density is reduced.

  • Convection is a form of heat transfer that involves the actual movement of particles of the medium.
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Learn more:

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7 0
3 years ago
g A box having mass 0,5kg is placed in front of a 20 cm compressed spring. When the spring released, box having mass m1, collide
vagabundo [1.1K]

Answer:Expression given below

Explanation:

Given mass of spring\left ( m_1\right )=0.5 kg

Compression in the spring\left ( x\right )=20 cm

Let the spring constant be K

Using Energy conservation

potential energy stored in spring =Kinetic energy of Block\left ( m_1\right )

\frac{1}{2}Kx^2=\frac{1}{2}m_1v^2

v=x\sqrt{\frac{k}{m_1}}

now conserving momentum

m_1v=\left ( m_1+m_2\right )v_0

v_0=\frac{m_1}{m_1+m_2}v

where v_0 is the final velocity

3 0
3 years ago
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dalvyx [7]
Water I believe sorry if wrong
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MAXImum [283]
Its acceleration is constant. 
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.Answer;

Using Fmax=qVB

F=(1.6*10^-19 C)(5.860*10^6 m/s)(1.38 T)

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